Introduction to eigenvalues and eigenvectors

WEBVTT
Kind: captions
Language: en

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For any transformation that maps
from Rn to Rn, we've done

00:00:06.870 --> 00:00:09.590
it implicitly, but it's been
interesting for us to find the

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vectors that essentially just
get scaled up by the

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transformations.

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So the vectors that have the
form-- the transformation of

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my vector is just equal
to some scaled-up

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version of a vector.

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And if this doesn't look
familiar, I can jog your

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memory a little bit.

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When we were looking for
basis vectors for the

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transformation--
let me draw it.

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This was from R2 to R2.

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So let me draw R2 right here.

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And let's say I had the
vector v1 was equal to

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the vector 1, 2.

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And we had the lines spanned
by that vector.

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We did this problem several
videos ago.

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And I had the transformation
that flipped across this line.

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So if we call that line l, T was
the transformation from R2

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to R2 that flipped vectors
across this line.

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So it flipped vectors
across l.

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So if you remember that
transformation, if I had some

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random vector that looked like
that, let's say that's x,

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that's vector x, then the
transformation of x looks

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something like this.

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It's just flipped across
that line.

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That was the transformation
of x.

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And if you remember that video,
we were looking for a

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change of basis that would allow
us to at least figure

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out the matrix for the
transformation, at least in an

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alternate basis.

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And then we could figure
out the matrix for the

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transformation in the
standard basis.

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And the basis we picked were
basis vectors that didn't get

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changed much by the
transformation, or ones that

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only got scaled by the
transformation.

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For example, when I took the
transformation of v1, it just

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equaled v1.

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Or we could say that the
transformation of v1 just

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equaled 1 times v1.

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So if you just follow this
little format that I set up

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here, lambda, in this
case, would be 1.

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And of course, the vector
in this case is v1.

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The transformation just
scaled up v1 by 1.

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In that same problem, we had
the other vector that

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we also looked at.

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It was the vector minus-- let's
say it's the vector v2,

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which is-- let's say
it's 2, minus 1.

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And then if you take the
transformation of it, since it

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was orthogonal to the
line, it just got

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flipped over like that.

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And that was a pretty
interesting vector force as

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well, because the transformation
of v2 in this

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situation is equal to what?

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Just minus v2.

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It's equal to minus v2.

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Or you could say that the
transformation of v2 is equal

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to minus 1 times v2.

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And these were interesting
vectors for us because when we

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defined a new basis with these
guys as the basis vector, it

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was very easy to figure out
our transformation matrix.

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And actually, that basis was
very easy to compute with.

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And we'll explore that a little
bit more in the future.

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But hopefully you realize that
these are interesting vectors.

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There was also the cases where
we had the planes spanned by

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some vectors.

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And then we had another vector
that was popping out of the

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plane like that.

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And we were transforming things
by taking the mirror

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image across this and we're
like, well in that

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transformation, these red
vectors don't change at all

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and this guy gets
flipped over.

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So maybe those would make
for good bases.

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Or those would make for
good basis vectors.

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And they did.

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So in general, we're always
interested with the vectors

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that just get scaled up
by a transformation.

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It's not going to be
all vectors, right?

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This vector that I drew here,
this vector x, it doesn't just

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get scaled up, it actually gets
changed, this direction

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gets changed.

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The vectors that get scaled up
might switch direct-- might go

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from this direction to that
direction, or maybe

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they go from that.

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Maybe that's x and then the
transformation of x might be a

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scaled up version of x.

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Maybe it's that.

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The actual, I guess, line that
they span will not change.

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And so that's what we're going
to concern ourselves with.

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These have a special name.

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And they have a special name and
I want to make this very

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clear because they're useful.

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It's not just some mathematical
game we're

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playing, although sometimes
we do fall into that trap.

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But they're actually useful.

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They're useful for defining
bases because in those bases

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it's easier to find
transformation matrices.

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They're more natural coordinate
systems. And

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oftentimes, the transformation
matrices in those bases are

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easier to compute with.

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And so these have
special names.

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Any vector that satisfies this
right here is called an

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eigenvector for the
transformation T.

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And the lambda, the multiple
that it becomes-- this is the

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eigenvalue associated with
that eigenvector.

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So in the example I just gave
where the transformation is

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flipping around this line,
v1, the vector 1, 2 is an

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eigenvector of our
transformation.

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So 1, 2 is an eigenvector.

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And it's corresponding
eigenvalue is 1.

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This guy is also an
eigenvector-- the

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vector 2, minus 1.

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He's also an eigenvector.

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A very fancy word, but all it
means is a vector that's just

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scaled up by a transformation.

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It doesn't get changed in any
more meaningful way than just

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the scaling factor.

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And it's corresponding
eigenvalue is minus 1.

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If this transformation--
I don't know what its

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transformation matrix is.

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I forgot what it was.

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We actually figured it
out a while ago.

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If this transformation matrix
can be represented as a matrix

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vector product-- and it should
be; it's a linear

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transformation-- then any
v that satisfies the

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transformation of-- I'll say
transformation of v is equal

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to lambda v, which also would
be-- you know, the

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transformation of [? v ?]

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would just be A times v.

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These are also called
eigenvectors of A, because A

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is just really the matrix
representation of the

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transformation.

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So in this case, this would be
an eigenvector of A, and this

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would be the eigenvalue
associated with the

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eigenvector.

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So if you give me a matrix that
represents some linear

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transformation.

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You can also figure
these things out.

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Now the next video we're
actually going to figure out a

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way to figure these
things out.

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But what I want you to
appreciate in this video is

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that it's easy to say,
oh, the vectors that

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don't get changed much.

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But I want you to understand
what that means.

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It literally just gets scaled up
or maybe they get reversed.

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Their direction or the
lines they span

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fundamentally don't change.

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And the reason why they're
interesting for us is, well,

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one of the reasons why they're
interesting for us is that

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they make for interesting basis
vectors-- basis vectors

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whose transformation matrices
are maybe computationally more

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simpler, or ones that make for
better coordinate systems.


Gram-Schmidt example with 3 basis vectors

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Kind: captions
Language: en

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Let's do one more Gram-Schmidt
example.

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So let's say I have the subspace
V that is spanned by

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the vectors-- let's say we're
dealing in R4, so the first

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vector is 0, 0, 1, 1.

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The second vector
is 0, 1, 1, 0.

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And then a third vector-- so
it's a three-dimensional

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subspace of R4-- it's 1, 1,
0, 0, just like that,

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three-dimensional
subspace of R4.

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And what we want to do, we want
to find an orthonormal

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basis for V.

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So we want to substitute these
guys with three other vectors

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that are orthogonal with
respect to each other

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and have length 1.

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So we do the same drill
we've done before.

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We can say-- let's call this
guy v1, this guy is v2, and

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let's call this guy v3.

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So the first thing we want to
do is replace v1-- and I'm

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just picking this guy at random
because he was the

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first guy on the
left-hand side.

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I want to replace v1 with an
orthogonal version of v1.

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So let me call u1 is equal to--
well, let me just find

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out the length the v1.

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I don't think I have to explain
too much of the theory

00:01:17.440 --> 00:01:17.960
at this point.

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I just want to show
another example.

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So the length of v1 is equal
to the square root of 0

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squared plus 0 squared plus 1
squared plus 1 squared, which

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equals the square root of 2.

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So let me define my new vector
u1 to be equal to 1 over the

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length of v1, 1 over the square
root of 2, times v1,

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times 0, 0, 1, 1.

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And just like that, the span
of v1, v2, v3, is the same

00:01:56.950 --> 00:02:01.040
thing is the span of
u1, v2, and v3.

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So this is my first thing
that I've normalized.

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So I can say that V is now
equal to the span of the

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vectors u1, v2, and v3.

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Because I can replace v1 with
this guy, because this guy is

00:02:17.900 --> 00:02:19.940
just a scaled-up version
of this guy.

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So I can definitely represent
him with him, so I can

00:02:23.000 --> 00:02:24.950
represent any linear combination
of these guys with

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any linear combination of
those guys right there.

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Now, we just did our
first vector.

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We just normalized this one.

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But we need to replace these
other vectors with vectors

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that are orthogonal to
this guy right here.

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So let's do v2 first. So let's
replace-- let's call it y2 is

00:02:42.850 --> 00:02:51.670
equal to v2 minus the projection
of v2 onto the

00:02:51.670 --> 00:02:56.530
space spanned by u1 or onto--
you know, I could call it c

00:02:56.530 --> 00:02:59.020
times u1, or in the past
videos, we called that

00:02:59.020 --> 00:03:03.820
subspace V1, but the space
spanned by u1.

00:03:03.820 --> 00:03:09.400
And that's just going to be
equal to y2 is equal to v2,

00:03:09.400 --> 00:03:18.210
which is 0, 1, 1, 0, minus-- v2
projected onto that space

00:03:18.210 --> 00:03:26.620
is just a dot product of v2, 0,
1, 1, 0, with the spanning

00:03:26.620 --> 00:03:29.030
vector of that space.

00:03:29.030 --> 00:03:31.020
And there's only one of them,
so we're only going to have

00:03:31.020 --> 00:03:36.240
one term like this with u1, so
dotted with 1 over the square

00:03:36.240 --> 00:03:44.940
root of 2 times 0, 0, 1, 1, and
then all of that times u1.

00:03:44.940 --> 00:03:51.340
So 1 over the square root of 2
times the vector 0, 0, 1, 1.

00:03:51.340 --> 00:03:55.330
And so this is going
to be equal to v2,

00:03:55.330 --> 00:03:58.750
which is 0, 1, 1, 0.

00:03:58.750 --> 00:04:00.890
The square root of 2, let's
factor them out.

00:04:00.890 --> 00:04:04.440
So then you just get-- or kind
of reassociate them out.

00:04:04.440 --> 00:04:06.603
So then you get this is 1 over
the square root of 2 times 1

00:04:06.603 --> 00:04:09.890
over the square root
of 2 is minus 1/2.

00:04:09.890 --> 00:04:12.640
You times-- what's the dot
product of these two guys?

00:04:12.640 --> 00:04:16.690
You get 0 times 0 plus 1 times
0, which is still 0, plus 1

00:04:16.690 --> 00:04:18.160
times 1 plus 0 times 0.

00:04:18.160 --> 00:04:22.930
So you're just going to have
times 1 times this out here:

00:04:22.930 --> 00:04:26.170
0, 0, 1, 1.

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I'll write that a little
bit neater.

00:04:27.735 --> 00:04:29.450
I'm getting careless.

00:04:29.450 --> 00:04:30.560
1, 1.

00:04:30.560 --> 00:04:39.750
So this is just going to be
equal to 0, 1, 1, 0 minus--

00:04:39.750 --> 00:04:41.380
1/2 times 0 is 0.

00:04:41.380 --> 00:04:42.640
1/2 times 0 is 0.

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Then I have two halves here.

00:04:46.730 --> 00:04:57.740
So y2 is equal to-- let's see,
0 minus 0 is 0, 1 minus 0 is

00:04:57.740 --> 00:05:08.940
1, 1 minus 1/2 is 1/2, and then
0 minus 1/2 is minus 1/2.

00:05:08.940 --> 00:05:19.460
So V, we can now write as the
span of u1, y2, and v3.

00:05:19.460 --> 00:05:20.520
And this is progress.

00:05:20.520 --> 00:05:25.350
u1 is orthogonal, y2-- sorry,
u1 is normalized.

00:05:25.350 --> 00:05:26.500
It has length 1.

00:05:26.500 --> 00:05:28.700
Y2 is orthogonal to it or
they're orthogonal with

00:05:28.700 --> 00:05:32.570
respect to each other, but y2
still has not been normalized.

00:05:32.570 --> 00:05:36.410
So let me replace y2 with a
normalized version of it.

00:05:36.410 --> 00:05:44.680
The length of y2 is equal to
the square root of 0 plus 1

00:05:44.680 --> 00:05:48.990
squared, which is 1, plus 1/2
squared, which is 1/4, plus

00:05:48.990 --> 00:05:54.360
minus 1/2 squared, which is
also 1/4, so plus 1/4.

00:05:54.360 --> 00:05:55.650
So this is 1 and 1/2.

00:05:55.650 --> 00:06:01.340
So it's equal to the
square root of 3/2.

00:06:01.340 --> 00:06:03.610
So let me define another
vector here.

00:06:03.610 --> 00:06:09.070
u2, which is equal to 1 over the
square root of 3/2, or we

00:06:09.070 --> 00:06:12.170
could say is the square root of
2/3, I'm just inverting it.

00:06:12.170 --> 00:06:14.540
It's 1 over the length of y2.

00:06:14.540 --> 00:06:16.620
So I'll just find the
reciprocal, so it's the square

00:06:16.620 --> 00:06:25.190
root 2 over 3 times y2, times
this guy right here, times 0,

00:06:25.190 --> 00:06:30.930
1, 1/2, and minus 1/2.

00:06:30.930 --> 00:06:34.200
And so this span is going to be
the same thing as the span

00:06:34.200 --> 00:06:39.660
of u1, u2, and v3.

00:06:39.660 --> 00:06:42.410
And there's our second
basis vector.

00:06:42.410 --> 00:06:43.400
And we're making a
lot of progress.

00:06:43.400 --> 00:06:45.870
These guys are orthogonal with
respect to each other.

00:06:45.870 --> 00:06:46.990
They both have length 1.

00:06:46.990 --> 00:06:49.750
We just have to do something
about v3.

00:06:49.750 --> 00:06:51.470
And we do it the same way.

00:06:51.470 --> 00:06:55.780
Let's find a vector that is
orthogonal to these guys, and

00:06:55.780 --> 00:06:58.370
if I sum that vector to some
linear combination of these

00:06:58.370 --> 00:07:00.650
guys, I'm going to get
v3, and I'm going to

00:07:00.650 --> 00:07:03.650
call that vector y3.

00:07:03.650 --> 00:07:13.230
y3 is equal to v3 minus the
projection of v3 onto the

00:07:13.230 --> 00:07:16.880
subspace spanned by u1 and u2.

00:07:16.880 --> 00:07:18.600
So I could call that
subspace-- let me

00:07:18.600 --> 00:07:19.150
just write it here.

00:07:19.150 --> 00:07:25.240
The span of u1 and u2, just
for notation, I'm going to

00:07:25.240 --> 00:07:26.670
call it v2.

00:07:26.670 --> 00:07:28.860
So it's v3, and actually,
I don't even have

00:07:28.860 --> 00:07:29.380
to write that .

00:07:29.380 --> 00:07:32.180
Minus the projection
of v3 onto that.

00:07:32.180 --> 00:07:33.620
And what's that going to be?

00:07:33.620 --> 00:07:43.150
That's going to be v3 dot u1
times u1, times the vector u1.

00:07:43.150 --> 00:07:49.080
And actually let me just--
plus v3 dot u2

00:07:49.080 --> 00:07:51.180
times the vector u2.

00:07:51.180 --> 00:07:54.420
Since this is an orthonormal
basis, the projection onto it,

00:07:54.420 --> 00:07:57.160
you just take the dot product
of v2 with each of their

00:07:57.160 --> 00:07:59.780
orthonormal basis vectors and
multiply them times the

00:07:59.780 --> 00:08:00.840
orthonormal basis vectors.

00:08:00.840 --> 00:08:02.180
We saw that several
videos ago.

00:08:02.180 --> 00:08:05.280
That's one of the neat things
about orthonormal bases.

00:08:05.280 --> 00:08:07.170
So what is this going
to be equal to?

00:08:07.170 --> 00:08:09.560
A little bit more computation
here.

00:08:09.560 --> 00:08:14.190
y3 is equal to v3, which
was up here.

00:08:14.190 --> 00:08:15.120
That's v3.

00:08:15.120 --> 00:08:17.520
v3 looks like this.

00:08:17.520 --> 00:08:24.820
It's 1, 1, 0, 0 minus
v3 dot u1.

00:08:24.820 --> 00:08:32.780
So this is minus v3,
1, 1, 0, 0, dot u1.

00:08:32.780 --> 00:08:43.309
So it's dot 1 over the square
root of 2 times 0, 0, 1, 1.

00:08:43.309 --> 00:08:48.230
That's u1-- so that's this part
right here-- times u1, so

00:08:48.230 --> 00:08:54.240
times 1 over the square root
of 2 times 0, 0, 1, 1.

00:08:54.240 --> 00:08:57.050
This piece right there is
this piece right there.

00:08:57.050 --> 00:08:59.230
And then we can distribute this
minus sign, so it's going

00:08:59.230 --> 00:09:00.160
to be plus.

00:09:00.160 --> 00:09:02.010
You know, we have a plus, but
there's this minus over here

00:09:02.010 --> 00:09:04.685
so we put a minus v3.

00:09:04.685 --> 00:09:06.540
Let me switch colors .

00:09:06.540 --> 00:09:15.680
Minus v3 , which is 1, 1 0, 0
dotted with u2, dotted with

00:09:15.680 --> 00:09:27.450
the square root of 2/3 times 0,
1, 1/2, minus 1/2 times u2,

00:09:27.450 --> 00:09:32.580
times the vector u2, times the
square root of 2/3, times the

00:09:32.580 --> 00:09:37.270
vector 0, 1, 1/2, minus 1/2.

00:09:37.270 --> 00:09:38.480
And what do we get?

00:09:38.480 --> 00:09:40.270
Let's calculate this.

00:09:40.270 --> 00:09:45.570
So we could take the-- so this
is going to be equal to the

00:09:45.570 --> 00:09:52.040
vector 1, 1, 0, 0, minus-- so
the 1 over the square root of

00:09:52.040 --> 00:09:53.390
2 and the 1 over the square
root of 2, multiply them.

00:09:53.390 --> 00:09:55.930
You're going to get a 1/2.

00:09:55.930 --> 00:09:58.140
And then when you take the dot
product of these two, 1 times

00:09:58.140 --> 00:10:01.650
0-- let's see, this is actually
all going to be, if

00:10:01.650 --> 00:10:05.430
you take the dot product of all
of these, then it actually

00:10:05.430 --> 00:10:06.920
gets 0, right?

00:10:06.920 --> 00:10:10.150
So this guy, v3, was actually
already orthogonal to u1.

00:10:12.920 --> 00:10:16.350
This will just go straight
to 0, which is nice.

00:10:16.350 --> 00:10:18.030
We don't have to have
a term right there.

00:10:18.030 --> 00:10:20.580
I took the dot product 1 times
0 plus 1 times 0 plus 0 times

00:10:20.580 --> 00:10:23.550
1 plus 0 times 1,
all gets zeroed.

00:10:23.550 --> 00:10:25.850
So this whole term drops out.

00:10:25.850 --> 00:10:29.120
We can ignore it, which makes
our computation simpler.

00:10:29.120 --> 00:10:32.790
And then over here we have minus
the square root of 2/3

00:10:32.790 --> 00:10:36.480
times the square root of 2/3
is just 2/3 times the dot

00:10:36.480 --> 00:10:37.860
product of these two guys.

00:10:37.860 --> 00:10:42.730
So that's 1 times 0, which is 0,
plus 1 times 1, which is 1,

00:10:42.730 --> 00:10:46.980
plus 0 times 1/2, which is 0,
plus 0 times minus 1/2, which

00:10:46.980 --> 00:10:54.630
is 0, so we just get a 1 there,
times the vector 0, 1,

00:10:54.630 --> 00:10:58.140
1/2, minus 1/2.

00:10:58.140 --> 00:10:59.730
And then what do we get?

00:10:59.730 --> 00:11:10.820
We get-- this is the home
stretch-- 1, 1, 0, 0 minus 2/3

00:11:10.820 --> 00:11:13.230
times all of these guys.

00:11:13.230 --> 00:11:14.940
So 2/3 time 0 is 0.

00:11:14.940 --> 00:11:17.340
2/3 times 1 is 2/3.

00:11:17.340 --> 00:11:23.870
2/3 times 1/2 is 1/3.

00:11:23.870 --> 00:11:29.120
And then 2/3 times minus
1/2 is minus 1/3.

00:11:29.120 --> 00:11:34.470
So then this is going to be
equal to 1 minus 0 is 1, 1

00:11:34.470 --> 00:11:40.290
minus 2/3 is 1/3, 0 minus 1/3 is
minus 1/3, and then 0 minus

00:11:40.290 --> 00:11:44.600
minus 1/3 is positive 1/3.

00:11:44.600 --> 00:11:47.790
So this vector y3 is orthogonal
to these two other

00:11:47.790 --> 00:11:50.800
vectors, which is nice, but it
still hasn't been normalized.

00:11:50.800 --> 00:11:52.770
So we finally have to normalize
this guy, and then

00:11:52.770 --> 00:11:53.210
we're done.

00:11:53.210 --> 00:11:54.450
Then we have an orthonormal
basis.

00:11:54.450 --> 00:11:58.690
We'll have u1, u2, and
now we'll find u3.

00:11:58.690 --> 00:12:05.020
So the length of my vector y--
actually, let's do something

00:12:05.020 --> 00:12:05.440
even better.

00:12:05.440 --> 00:12:07.360
It'll simplify things
a little bit.

00:12:07.360 --> 00:12:09.440
Instead of a writing
y this way, I

00:12:09.440 --> 00:12:10.650
could scale up y, right?

00:12:10.650 --> 00:12:13.150
All I want is a vector that's
orthogonal to the other two

00:12:13.150 --> 00:12:14.870
that still spans
the same space.

00:12:14.870 --> 00:12:16.650
So I can scale this guy up.

00:12:16.650 --> 00:12:21.740
So I could say, I don't know,
let me call it y3-- let me

00:12:21.740 --> 00:12:23.400
call it y3 prime.

00:12:23.400 --> 00:12:24.900
And I'm just doing this to
ease the computation.

00:12:24.900 --> 00:12:28.250
I could just scale this guy
up, multiply him by 3.

00:12:28.250 --> 00:12:29.910
So what do I get?

00:12:29.910 --> 00:12:31.220
I probably should have done
it some of the other ones.

00:12:31.220 --> 00:12:36.870
3, 1, minus 1, and 1.

00:12:36.870 --> 00:12:40.370
And so I can replace y3 with
this guy, and then I can just

00:12:40.370 --> 00:12:41.440
normalize this guy.

00:12:41.440 --> 00:12:42.960
It'll be a little bit easier.

00:12:42.960 --> 00:12:48.540
So the length of y3 prime that I
just defined is equal to the

00:12:48.540 --> 00:12:55.750
square root of 3 squared, which
is 9, plus 1 squared

00:12:55.750 --> 00:12:58.350
plus minus 1 squared plus 1
squared, which is equal to the

00:12:58.350 --> 00:13:01.500
square root of 12,
which is what?

00:13:01.500 --> 00:13:03.360
That's two square roots of 3.

00:13:03.360 --> 00:13:06.360
That is equal to 2 square
roots of 3, right?

00:13:06.360 --> 00:13:09.260
Square root of 4 times the
square root of 3, which is two

00:13:09.260 --> 00:13:10.450
square roots of 3.

00:13:10.450 --> 00:13:16.630
So now I can to find u3 as equal
to y3 times 1 over the

00:13:16.630 --> 00:13:20.790
length of y3, so it's equal to
1 over two square roots of 3

00:13:20.790 --> 00:13:25.720
times the vector 3,
1, minus 1, and 1.

00:13:25.720 --> 00:13:26.970
And then we're done.

00:13:26.970 --> 00:13:30.020
If we have a basis, an
orthonormal basis would be

00:13:30.020 --> 00:13:33.520
this guy-- let me take the other
ones down here-- and

00:13:33.520 --> 00:13:35.590
these guys.

00:13:35.590 --> 00:13:41.470
All of these form-- let me bring
it all the way down.

00:13:41.470 --> 00:13:44.960
If I have a collection of these
three vectors, I now

00:13:44.960 --> 00:13:51.180
have an orthonormal basis for
V, these three right there.

00:13:51.180 --> 00:13:54.390
That set is an orthonormal basis
for my original subspace

00:13:54.390 --> 00:13:57.010
V that I started off with.


Gram-Schmidt process example

WEBVTT
Kind: captions
Language: en

00:00:00.660 --> 00:00:03.710
We came up with a process for
generating an orthonormal

00:00:03.710 --> 00:00:06.515
basis in the last video, and
it wasn't a new discovery.

00:00:06.515 --> 00:00:08.650
It's called the Gram-Schmidt
process.

00:00:08.650 --> 00:00:11.760
But let's apply that now to
some real examples, and

00:00:11.760 --> 00:00:13.590
hopefully, we'll see that it's
a lot more concrete than it

00:00:13.590 --> 00:00:14.780
might have looked in
the last video.

00:00:14.780 --> 00:00:24.040
Let's say I have the plane x1
plus x2 plus x3 is equal to 0.

00:00:24.040 --> 00:00:26.200
This is a plane in R3.

00:00:26.200 --> 00:00:32.549
So let's just say that subspace
V is equal to the

00:00:32.549 --> 00:00:41.370
plane defined by this guy right
here. x1 plus x2 plus x3

00:00:41.370 --> 00:00:42.390
is equal to 0.

00:00:42.390 --> 00:00:44.760
All of the vectors in the
subspace, if you take their

00:00:44.760 --> 00:00:47.190
entries and you add them up,
you're going to get 0.

00:00:47.190 --> 00:00:50.001
So first we need just any basis
for v, so let's see if

00:00:50.001 --> 00:00:51.510
we can come up with that.

00:00:51.510 --> 00:00:53.750
So if we subtract x2 and x3
from both sides of this

00:00:53.750 --> 00:00:57.060
equation, we know that x1
is going to be equal to

00:00:57.060 --> 00:01:00.440
minus x2 minus x3.

00:01:00.440 --> 00:01:04.680
Or we could say that our
subspace V is equal to the set

00:01:04.680 --> 00:01:12.240
of all of the vectors in R3--
x1, x2 and x3-- that satisfy

00:01:12.240 --> 00:01:18.120
the equation, let's say,
minus-- well, let me

00:01:18.120 --> 00:01:18.700
write it this way.

00:01:18.700 --> 00:01:26.770
Let's say that x2 is equal to
c1, and x3 is equal to c2.

00:01:26.770 --> 00:01:33.980
Then this equation would be x1
is equal to minus c1 minus c2.

00:01:33.980 --> 00:01:36.980
So if we write it that way, then
the subspace V is a set

00:01:36.980 --> 00:01:42.860
of all of the vectors in R3
such that c1 times some

00:01:42.860 --> 00:01:47.700
vector-- let me write it this
way-- c1 times-- let me write

00:01:47.700 --> 00:01:51.880
it this way-- plus c2 times some
other vector, where c1

00:01:51.880 --> 00:01:56.530
and c2 are any real numbers,
so c1 and c2 are a

00:01:56.530 --> 00:01:58.890
member of the reals.

00:01:58.890 --> 00:02:03.600
And so what is x1? x1 is equal
to minus c1 minus c2.

00:02:03.600 --> 00:02:09.139
So x1 is equal to minus
1 c1, minus 1 c2.

00:02:09.139 --> 00:02:11.190
x2 is just equal to c1.

00:02:11.190 --> 00:02:15.550
So x2 is equal to 1 times
c1 plus 0 times c2.

00:02:15.550 --> 00:02:21.650
And then x3 is equal to c2, or
0 times c1 plus 1 times c2.

00:02:21.650 --> 00:02:25.040
So V is essentially the span of
these two vectors, all of

00:02:25.040 --> 00:02:26.940
the linear combinations
of these two vectors.

00:02:26.940 --> 00:02:28.520
That would represent
that plane.

00:02:28.520 --> 00:02:29.490
So let me write it like this.

00:02:29.490 --> 00:02:39.580
So V is equal to the span of the
vectors minus 1, 1, 0, and

00:02:39.580 --> 00:02:44.480
the vector minus 1, 0, 1.

00:02:44.480 --> 00:02:46.350
And you can see that
these are linearly

00:02:46.350 --> 00:02:48.750
independent right here.

00:02:48.750 --> 00:02:52.260
Obviously, there's no linear
combination of this guy that

00:02:52.260 --> 00:02:54.090
can give you a 1 over here,
and there's no linear

00:02:54.090 --> 00:02:55.290
combination of this
guy that'll give

00:02:55.290 --> 00:02:56.950
you a 1 right there.

00:02:56.950 --> 00:02:59.010
So this is what V is.

00:02:59.010 --> 00:03:01.920
But what we want, the whole
reason why I'm making this

00:03:01.920 --> 00:03:04.870
video, is to find an orthonormal
basis for V.

00:03:04.870 --> 00:03:06.990
This is just a basis.

00:03:06.990 --> 00:03:09.040
These guys right here are
just a basis for V.

00:03:09.040 --> 00:03:11.480
Let's find an orthonormal
basis.

00:03:11.480 --> 00:03:14.860
Let's call this vector up here,
let's call that v1, and

00:03:14.860 --> 00:03:18.370
let's call this vector
right here v2.

00:03:18.370 --> 00:03:23.480
So if we wanted to find an
orthonormal basis for the span

00:03:23.480 --> 00:03:25.520
of v1-- let me write
this down.

00:03:25.520 --> 00:03:31.150
Let me define some subspace V1
is equal to the span of just

00:03:31.150 --> 00:03:32.910
my vector v1.

00:03:32.910 --> 00:03:35.740
Well, we saw in the last video,
if we just divide v1 by

00:03:35.740 --> 00:03:40.640
its length, then the span of
that vector is going to be a

00:03:40.640 --> 00:03:43.280
unit vector, and it's going to
be the same thing as the

00:03:43.280 --> 00:03:44.050
subspace V1.

00:03:44.050 --> 00:03:45.970
It's this line in R3.

00:03:45.970 --> 00:03:46.660
So let's do that.

00:03:46.660 --> 00:03:49.110
What is the length of v1?

00:03:49.110 --> 00:03:55.870
The length of v1 is equal to
the square root of minus 1

00:03:55.870 --> 00:04:00.030
squared, which is 1, plus 1
squared, which is 1, plus 0

00:04:00.030 --> 00:04:03.650
squared, which is 0, so it's
equal to the square root of 2.

00:04:03.650 --> 00:04:10.030
So let's define some vector u1
is equal to 1 divided by the

00:04:10.030 --> 00:04:14.010
length of v1, so 1 over the
square root of 2 times v1,

00:04:14.010 --> 00:04:18.720
times minus 1, 1, 0.

00:04:18.720 --> 00:04:21.029
Then the span of v1 is
just the same thing

00:04:21.029 --> 00:04:25.530
as the span of u1.

00:04:25.530 --> 00:04:27.620
And so this would be an
orthonormal basis.

00:04:27.620 --> 00:04:29.930
Just this vector right here
would be an orthonormal basis

00:04:29.930 --> 00:04:31.600
for just the span of v1.

00:04:31.600 --> 00:04:33.880
But we don't want just the span
of v1, we want the span

00:04:33.880 --> 00:04:35.910
of v1 and v2.

00:04:35.910 --> 00:04:38.230
Let me just draw it.

00:04:38.230 --> 00:04:41.210
So right now, we have a basis,
if I just do u1.

00:04:41.210 --> 00:04:42.690
I'm not going to actually draw
what this looks like.

00:04:42.690 --> 00:04:45.900
Maybe it looks something like
this, and its span is this

00:04:45.900 --> 00:04:47.150
entire line in R3.

00:04:50.160 --> 00:04:52.860
The span of just one vector in
Rn is just going to be all the

00:04:52.860 --> 00:04:55.600
scalar multiples of it
or a line in Rn.

00:04:55.600 --> 00:04:58.310
So this right here is
the subspace V1.

00:04:58.310 --> 00:05:01.700
Now, we have a v2 here, which
is linearly independent from

00:05:01.700 --> 00:05:03.690
this guy, which means it's
linearly independent from this

00:05:03.690 --> 00:05:06.360
guy, because he's just a scaled
version of this guy.

00:05:06.360 --> 00:05:10.480
So v2 is going to
look like that.

00:05:10.480 --> 00:05:12.510
That is v2 right there.

00:05:12.510 --> 00:05:15.070
This, of course, was our u1.

00:05:15.070 --> 00:05:21.350
And what we want to do is we
want to find a subspace V.

00:05:21.350 --> 00:05:23.690
I'll call it V2 for now.

00:05:23.690 --> 00:05:31.530
V2 is equal to the span of v1
and v2, which is the same

00:05:31.530 --> 00:05:35.720
thing as the span-- anything
that's spanned by v1 is also

00:05:35.720 --> 00:05:40.860
spanned by u1-- the
span of u1 and v2.

00:05:40.860 --> 00:05:43.090
So we want to see everything
that could be generated by

00:05:43.090 --> 00:05:45.480
linear combinations
of u1 and v2.

00:05:45.480 --> 00:05:48.300
And obviously, this thing right
here is our plane that

00:05:48.300 --> 00:05:48.930
we're talking about.

00:05:48.930 --> 00:05:53.330
The span of these two guys,
that is the whole subspace

00:05:53.330 --> 00:05:54.540
that we're talking about
in this problem.

00:05:54.540 --> 00:05:56.630
So that is equal to V.

00:05:56.630 --> 00:05:59.550
So once we find this, if we find
an orthonormal version

00:05:59.550 --> 00:06:01.215
for this span, we're done.

00:06:01.215 --> 00:06:02.900
So how can we do that?

00:06:02.900 --> 00:06:06.140
Well, if I can find a vector
that's orthogonal to all of

00:06:06.140 --> 00:06:09.380
the linear combinations of this,
that if I add up some

00:06:09.380 --> 00:06:12.040
linear combination of this to
that vector, I can get v2, I

00:06:12.040 --> 00:06:14.480
can replace v2 with
that vector.

00:06:14.480 --> 00:06:19.890
So we can call that vector
right there y2, right?

00:06:19.890 --> 00:06:22.680
If I can determine a y2, this
y2 is clearly orthogonal to

00:06:22.680 --> 00:06:27.490
everything over here, and I can
take some vector in v1, in

00:06:27.490 --> 00:06:30.220
this line, and add it to y2,
and I can get to v2.

00:06:30.220 --> 00:06:33.390
So combinations of these guys
are just as good as v2.

00:06:33.390 --> 00:06:41.180
So this is going to be equal
to the span of u1 and y2.

00:06:41.180 --> 00:06:43.330
Now what is y2 equal to?

00:06:43.330 --> 00:06:44.930
Well, we saw in the last
video, this is just a

00:06:44.930 --> 00:06:47.630
projection of v2.

00:06:47.630 --> 00:06:51.640
This vector right here is the
projection of v2 onto the

00:06:51.640 --> 00:06:53.370
subspace V1.

00:06:53.370 --> 00:06:54.870
And how do we figure out
what that-- and then

00:06:54.870 --> 00:06:55.910
what would y2 be?

00:06:55.910 --> 00:06:58.110
y2 would be v2 minus that.

00:06:58.110 --> 00:07:06.840
So y2 is equal to v2 minus the
projection of v2 onto v1.

00:07:06.840 --> 00:07:09.430
Or if, we were to actually write
it out, what is that

00:07:09.430 --> 00:07:11.060
going to be equal to?

00:07:11.060 --> 00:07:15.870
So it's going to be equal to--
v2 is this vector right here.

00:07:15.870 --> 00:07:19.050
So it's minus 1, 0, 2.

00:07:19.050 --> 00:07:21.370
That is v2.

00:07:21.370 --> 00:07:24.820
v2 minus the projection
of v2 onto v1.

00:07:24.820 --> 00:07:28.190
Well the projection of the
vector v2 onto the subspace V1

00:07:28.190 --> 00:07:35.780
is just v2, minus 1, 0, 0,
dotted with the orthonormal

00:07:35.780 --> 00:07:37.690
basis for v1.

00:07:37.690 --> 00:07:40.655
The orthonormal basis
for v1 is just u1.

00:07:40.655 --> 00:07:43.620
And we solved for u1 up here,
so that's going to be that

00:07:43.620 --> 00:07:48.890
dotted with 1 over the square
root of 2 times-- let me do

00:07:48.890 --> 00:07:50.450
that in yellow, actually,
just so you can see

00:07:50.450 --> 00:07:51.490
that this is u1.

00:07:51.490 --> 00:07:54.690
So dotting it with u1, so
dotting it with 1 over the

00:07:54.690 --> 00:08:00.780
square root of 2 times minus 1,
1, 0-- I like leaving the 1

00:08:00.780 --> 00:08:02.670
over the square root of 2 out of
there, just to keep things

00:08:02.670 --> 00:08:05.960
simple-- all of that divided
by-- actually,

00:08:05.960 --> 00:08:07.720
not divided by anything.

00:08:07.720 --> 00:08:10.680
Because we if we were doing a
projection of a line, it would

00:08:10.680 --> 00:08:15.080
be divided by the dot product of
the orthonormal basis with

00:08:15.080 --> 00:08:17.360
itself, but its length is 1, so
we don't have that, and we

00:08:17.360 --> 00:08:18.150
saw that before.

00:08:18.150 --> 00:08:22.310
Actually let me-- let me write
this a little bit.

00:08:22.310 --> 00:08:23.700
I'll just move it down.

00:08:23.700 --> 00:08:26.630
Let me see if I can move it.

00:08:26.630 --> 00:08:31.370
It's just equal to
that guy, right?

00:08:31.370 --> 00:08:32.850
Let me make the numbers clear.

00:08:32.850 --> 00:08:39.130
This is v2 minus the projection
of v2 onto the

00:08:39.130 --> 00:08:39.880
subspace 1.

00:08:39.880 --> 00:08:46.840
So that's just v2 dotted with
my orthonormal basis for v1,

00:08:46.840 --> 00:08:48.680
my first vector in my
orthonormal basis.

00:08:48.680 --> 00:08:50.830
There's only one, so I'm only
going to have one term here,

00:08:50.830 --> 00:08:53.960
and then all of that times
my orthonormal

00:08:53.960 --> 00:08:55.950
basis vector for v1.

00:08:55.950 --> 00:08:59.050
So 1 over the square
root of 2 times the

00:08:59.050 --> 00:09:03.070
vector minus 1, 1, 0.

00:09:03.070 --> 00:09:04.860
Now this looks really fancy.

00:09:04.860 --> 00:09:08.420
This right here is our
orthonormal basis for the

00:09:08.420 --> 00:09:12.020
subspace V1, but what does
this simplify to?

00:09:12.020 --> 00:09:15.450
So this is going to be equal
to-- remember, this right

00:09:15.450 --> 00:09:18.120
here, this piece right
there, that's the

00:09:18.120 --> 00:09:21.970
projection onto v1 of v2.

00:09:21.970 --> 00:09:24.230
That's what that was
right there.

00:09:24.230 --> 00:09:29.880
So this is going to be equal to
the vector minus 1, 0, 1,

00:09:29.880 --> 00:09:33.140
minus-- now, I can take the 1
over the square root of 2 on

00:09:33.140 --> 00:09:33.620
the outside.

00:09:33.620 --> 00:09:35.470
Actually, I can take both of
these onto the outside.

00:09:35.470 --> 00:09:37.385
So 1 over the square root of 2
times 1 over the square root

00:09:37.385 --> 00:09:41.670
of 2, it's just going to
be 1 over 2, right?

00:09:41.670 --> 00:09:47.270
So this is going to be minus
1/2 times these guys dotted

00:09:47.270 --> 00:09:48.580
with each other.

00:09:48.580 --> 00:09:49.430
Let me just write it this way.

00:09:49.430 --> 00:09:51.600
So what is these guys dotted
with each other?

00:09:51.600 --> 00:09:54.070
It's just going to
be a number.

00:09:54.070 --> 00:09:58.760
Minus 1 times minus 1 is 1,
plus 0 times 1, so plus 0,

00:09:58.760 --> 00:10:01.440
plus 1 times 0, so plus 0.

00:10:01.440 --> 00:10:04.440
All of that times-- well, we
already used this part of it,

00:10:04.440 --> 00:10:07.810
so we just have this part
left over-- times minus

00:10:07.810 --> 00:10:09.700
1, 1, and the 0.

00:10:09.700 --> 00:10:13.420
This was the dot product, and
we took the two scaling

00:10:13.420 --> 00:10:14.110
factors out.

00:10:14.110 --> 00:10:16.830
When you multiply them you got
1/2, so this is just going to

00:10:16.830 --> 00:10:19.440
be a 1, which simplifies
things.

00:10:19.440 --> 00:10:25.680
So this is going to be equal to
the vector minus 1, 0, 1,

00:10:25.680 --> 00:10:32.350
minus 1/2 times this, or we
could just write-- 1/2 times

00:10:32.350 --> 00:10:35.150
minus 1 is minus 1/2.

00:10:35.150 --> 00:10:38.190
We have 1/2 and then
we have 0.

00:10:38.190 --> 00:10:41.120
And so this is going to
be equal to minus

00:10:41.120 --> 00:10:44.010
1, minus minus 1/2.

00:10:44.010 --> 00:10:48.220
That's plus 1/2, so that's going
to be just minus 1/2.

00:10:48.220 --> 00:10:51.050
0 minus 1/2 is minus 1/2.

00:10:51.050 --> 00:10:54.120
And then 1 minus 0 is just 1.

00:10:54.120 --> 00:10:56.790
So this right here
is our vector y2.

00:10:59.610 --> 00:11:04.320
And if you combined u1 right
here and y2, we are spanning

00:11:04.320 --> 00:11:05.930
our subspace V.

00:11:05.930 --> 00:11:08.000
But we don't have an orthonormal
basis yet.

00:11:08.000 --> 00:11:10.490
These guys are orthogonal with
respect to each other, but

00:11:10.490 --> 00:11:12.840
this guy does not have
length 1 yet.

00:11:12.840 --> 00:11:15.500
So to make it equal to length
1, let's replace him.

00:11:15.500 --> 00:11:22.700
Let's define another vector u2
that's equal to 1 over the

00:11:22.700 --> 00:11:29.820
length of y2 times y2.

00:11:29.820 --> 00:11:31.320
So what is the length of y2?

00:11:31.320 --> 00:11:38.290
The length of y2 is equal to the
square root of minus 1/2

00:11:38.290 --> 00:11:45.160
squared is 1/4, plus
1 squared.

00:11:45.160 --> 00:11:48.930
So it's the square root
of 1 and 1/2, or 3/2.

00:11:48.930 --> 00:11:54.060
So it's equal to the square
root of 3/2, right?

00:11:54.060 --> 00:11:59.250
Yeah, this is 1/2 plus 1 is
1 and 1/2, which is 3/2.

00:11:59.250 --> 00:12:03.080
So it's equal to the
square root of 3/2.

00:12:03.080 --> 00:12:07.806
So if I defined u2, u2 is equal
to 1 over the square

00:12:07.806 --> 00:12:11.120
root of 3/2, or that's the same
thing as the square root

00:12:11.120 --> 00:12:18.060
of 2/3 times y2, which is this
guy right here, minus 1/2,

00:12:18.060 --> 00:12:20.470
minus 1/2, and 1.

00:12:20.470 --> 00:12:23.420
And I already had defined
u1 up here.

00:12:23.420 --> 00:12:26.680
u1 was right up here.

00:12:26.680 --> 00:12:28.540
Let me copy and paste.

00:12:28.540 --> 00:12:32.180
Actually, I think I can
just move it down.

00:12:36.600 --> 00:12:38.140
So I found u1 right here.

00:12:38.140 --> 00:12:44.930
We now have two vectors that are
orthogonal with respect to

00:12:44.930 --> 00:12:45.950
each other.

00:12:45.950 --> 00:12:50.870
So if I have the set of
u1 and u2, these guys

00:12:50.870 --> 00:12:52.410
both have length 1.

00:12:52.410 --> 00:12:54.620
They are orthogonal with respect
to each other, and

00:12:54.620 --> 00:12:56.410
they span V.

00:12:56.410 --> 00:13:03.030
So this is an orthonormal basis
for the plane that we

00:13:03.030 --> 00:13:06.340
started this video
out with: for V.

00:13:06.340 --> 00:13:07.030
And we're done.

00:13:07.030 --> 00:13:10.650
We have done the Gram-Schmidt
process.

00:13:10.650 --> 00:13:13.390
These are our new orthonormal
basis vectors.


The Gram-Schmidt process

WEBVTT
Kind: captions
Language: en

00:00:00.760 --> 00:00:04.780
Let's say I have a set of
linearly independent vectors,

00:00:04.780 --> 00:00:16.460
V1, V2, all the way to Vk,
that are a basis for V.

00:00:16.460 --> 00:00:17.880
We've seen this many
times before.

00:00:17.880 --> 00:00:18.450
That's nice.

00:00:18.450 --> 00:00:20.570
It's a basis, but we've learned
over the last few

00:00:20.570 --> 00:00:22.900
videos it would be even
better to have an

00:00:22.900 --> 00:00:30.280
orthonormal basis for V.

00:00:32.950 --> 00:00:35.500
What we're going to address in
this video is, is there some

00:00:35.500 --> 00:00:40.090
way that we can construct an
orthonormal basis for V, given

00:00:40.090 --> 00:00:42.700
that we already have this basis,
which I'm assuming

00:00:42.700 --> 00:00:43.630
isn't orthonormal.

00:00:43.630 --> 00:00:45.700
It'll work whether or not
it's orthonormal.

00:00:45.700 --> 00:00:48.690
It'll just generate another
orthonormal basis.

00:00:48.690 --> 00:00:51.680
But can we somehow, just given
any basis, generate an

00:00:51.680 --> 00:00:55.320
orthonormal basis for V, and
then be able to benefit from

00:00:55.320 --> 00:00:57.820
all of the properties of having
an orthonormal basis?

00:00:57.820 --> 00:01:02.330
Let's see if we can make
some headway here.

00:01:02.330 --> 00:01:05.000
Let's think about it
for simple cases.

00:01:05.000 --> 00:01:09.080
Let's say that I have a
one-dimensional subspace.

00:01:09.080 --> 00:01:12.540
Let's call that V1, for it's
a one-dimensional subspace.

00:01:12.540 --> 00:01:20.870
I'm going to say it's the span
of just this vector V1.

00:01:20.870 --> 00:01:26.210
Now, or you could say the vector
V1 is the basis for the

00:01:26.210 --> 00:01:27.770
subspace V1.

00:01:27.770 --> 00:01:31.290
If I had this simple case, how
can I ensure that this is

00:01:31.290 --> 00:01:33.030
orthonormal?

00:01:33.030 --> 00:01:36.266
What I could do is I could
define some vector,

00:01:36.266 --> 00:01:37.790
let's call it U1.

00:01:42.560 --> 00:01:44.850
Obviously, this is orthogonal
to all of the other guys.

00:01:44.850 --> 00:01:46.360
There aren't any other
guys here.

00:01:46.360 --> 00:01:47.600
[INAUDIBLE]

00:01:47.600 --> 00:01:50.520
vectors there, but there aren't
any other members of

00:01:50.520 --> 00:01:53.270
this set, so it's orthogonal
to everything else because

00:01:53.270 --> 00:01:54.660
there isn't anything else.

00:01:54.660 --> 00:01:58.470
And then to make its length
equal one, you can take this

00:01:58.470 --> 00:02:01.120
vector and divide it
by its length.

00:02:01.120 --> 00:02:07.810
So if we define some vector U1
to be equal to V1 divided by

00:02:07.810 --> 00:02:14.330
the length of V1, then
the length of U1

00:02:14.330 --> 00:02:15.050
is going to be what?

00:02:15.050 --> 00:02:22.500
It's going to be the length
of V1 over the

00:02:22.500 --> 00:02:24.340
length of V1 like this.

00:02:24.340 --> 00:02:26.360
This is just a constant
right here.

00:02:26.360 --> 00:02:29.570
This is going to be 1 over
the length of V1.

00:02:29.570 --> 00:02:34.090
That could be 5, times length
of V1, which is

00:02:34.090 --> 00:02:35.070
just equal to 1.

00:02:35.070 --> 00:02:36.890
This one will have
a length of 1.

00:02:36.890 --> 00:02:45.200
So just like that, if we have
the set of just U1, we could

00:02:45.200 --> 00:02:57.740
say that just the set of U1 is
an orthonormal basis for V1,

00:02:57.740 --> 00:02:59.430
for the subspace V1.

00:02:59.430 --> 00:03:01.690
Now, that was trivially easy.

00:03:01.690 --> 00:03:04.110
If k happens to be
1, we're done.

00:03:04.110 --> 00:03:05.440
That was super easy.

00:03:05.440 --> 00:03:07.040
We just have to divide
by our length.

00:03:07.040 --> 00:03:08.990
You have to just essentially
just normalize this vector,

00:03:08.990 --> 00:03:11.060
and you have an orthonormal
basis because there's nothing

00:03:11.060 --> 00:03:12.700
else to be orthogonal with.

00:03:12.700 --> 00:03:14.570
Well let's complicated the
issue a little bit.

00:03:14.570 --> 00:03:17.200
Let's say we go to
two dimensions.

00:03:17.200 --> 00:03:20.900
Let's say we have
a subspace V2.

00:03:20.900 --> 00:03:22.520
That's the span of, let's
say it's the first

00:03:22.520 --> 00:03:24.190
two of these vectors.

00:03:24.190 --> 00:03:34.810
It's the span of V1
and the vector V2.

00:03:34.810 --> 00:03:40.580
Now, we know that V1, V1 can
easily be represented as-- or,

00:03:40.580 --> 00:03:43.020
V1 is a linear combination
of U1.

00:03:43.020 --> 00:03:43.800
How do I know that?

00:03:43.800 --> 00:03:45.800
Well, I can just multiply
both sides of this times

00:03:45.800 --> 00:03:47.030
the length of V1.

00:03:47.030 --> 00:03:54.240
We have U1 times the length
of V1 is equal to V1.

00:03:54.240 --> 00:03:56.440
So we could say this
is the same thing.

00:03:56.440 --> 00:04:04.320
This is equivalent to saying the
span of the vector U1 and

00:04:04.320 --> 00:04:08.780
the vector V2, where U1 is the
vector we got up here.

00:04:08.780 --> 00:04:09.670
How can I say this?

00:04:09.670 --> 00:04:12.160
Because anything that can be a
linear combination of these

00:04:12.160 --> 00:04:14.560
guys can be a linear combination
of those guys.

00:04:14.560 --> 00:04:17.230
Because where ever you had a
V1, you can replace the V1

00:04:17.230 --> 00:04:20.269
with the linear combination
of U1 that gives you V1.

00:04:20.269 --> 00:04:22.430
So you just multiply U1 times
that scalar and you get it.

00:04:22.430 --> 00:04:24.140
I think you get the point.

00:04:24.140 --> 00:04:27.470
But how can we ensure that this
is an orthonormal set?

00:04:27.470 --> 00:04:28.110
What do we do?

00:04:28.110 --> 00:04:29.100
Let's graph it.

00:04:29.100 --> 00:04:30.740
This is going to be
a plane in Rn.

00:04:30.740 --> 00:04:33.320
So let's just make our
chalkboard here, or our video

00:04:33.320 --> 00:04:34.430
board here, the plane.

00:04:34.430 --> 00:04:36.950
So we have U1, which
is a unit vector.

00:04:36.950 --> 00:04:38.150
It has length 1.

00:04:38.150 --> 00:04:39.400
So that is U1.

00:04:42.950 --> 00:04:46.000
V1 and V2 are linearly
independent, that's by

00:04:46.000 --> 00:04:47.120
definition of a basis.

00:04:47.120 --> 00:04:51.250
So you can't represent V2 as
a linear multiple or linear

00:04:51.250 --> 00:04:53.300
combination of V1.

00:04:53.300 --> 00:04:56.600
Likewise, you can't represent V2
as a linear combination of

00:04:56.600 --> 00:04:59.750
U1 because U1 is a linear
combination of U1.

00:04:59.750 --> 00:05:03.130
So, V1 is not going to be on
the line spanned by U1.

00:05:03.130 --> 00:05:04.390
So I could actually draw that.

00:05:04.390 --> 00:05:07.980
The line spanned by U1
is just like this.

00:05:07.980 --> 00:05:10.290
This is the line
spanned by U1.

00:05:10.290 --> 00:05:13.350
Let me draw it a little
bit better than that.

00:05:13.350 --> 00:05:14.330
I don't want to make
it too dark.

00:05:14.330 --> 00:05:17.500
So this is the line.

00:05:17.500 --> 00:05:19.570
I'll do one last try.

00:05:19.570 --> 00:05:22.725
The line spanned by U1
looks like this.

00:05:22.725 --> 00:05:24.730
It looks like that,
it's just a line.

00:05:24.730 --> 00:05:29.610
And that is the subspace
V1, right?

00:05:29.610 --> 00:05:31.590
With the span of U1.

00:05:31.590 --> 00:05:35.390
So this is equal to
the span of U1.

00:05:35.390 --> 00:05:38.480
All we did is normalize V1
right there to get to U1.

00:05:38.480 --> 00:05:41.280
So the span of V1 is the
same span of U1.

00:05:41.280 --> 00:05:42.960
That is this subspace.

00:05:42.960 --> 00:05:45.040
That's this line in Rn.

00:05:45.040 --> 00:05:47.070
We have vector V2,
which is linearly

00:05:47.070 --> 00:05:49.120
independent from V1 and U1.

00:05:49.120 --> 00:05:53.960
V2 would look, let's just
say something like that.

00:05:53.960 --> 00:05:55.800
V2.

00:05:55.800 --> 00:06:00.130
Now, what we want to do is
replace V2 with another vector

00:06:00.130 --> 00:06:03.550
that is definitely orthogonal
to this, and will still be

00:06:03.550 --> 00:06:07.570
able to construct V2 with some
combination of this and our

00:06:07.570 --> 00:06:08.840
new vector.

00:06:08.840 --> 00:06:12.780
Well, the most obvious vector
would be some vector that is

00:06:12.780 --> 00:06:17.540
orthogonal to V1.

00:06:17.540 --> 00:06:20.820
So it's orthogonal, so it's
a member of the orthogonal

00:06:20.820 --> 00:06:22.070
complement of V1.

00:06:26.560 --> 00:06:29.200
If you look at it just visually
here, if I add some

00:06:29.200 --> 00:06:33.740
member of V1 to that member of
the orthogonal complement of

00:06:33.740 --> 00:06:35.210
V1, I'm going to get V2.

00:06:35.210 --> 00:06:37.125
In fact, we've seen that
multiple times.

00:06:40.420 --> 00:06:44.540
We know that any vector in
Rn, let's say V2, can be

00:06:44.540 --> 00:06:46.880
represented as a sum
of two vectors.

00:06:46.880 --> 00:06:58.740
I'll call them x and y, where x
is a member of V1 and y is a

00:06:58.740 --> 00:07:01.510
member of the orthogonal
complement of V1.

00:07:01.510 --> 00:07:03.700
We've seen that multiple
times.

00:07:03.700 --> 00:07:07.890
Now what was this thing
by definition?

00:07:07.890 --> 00:07:08.890
We're looking for this thing.

00:07:08.890 --> 00:07:11.950
This is the x.

00:07:11.950 --> 00:07:14.590
And this is the y.

00:07:14.590 --> 00:07:18.230
We're looking for the y, because
if we can find this

00:07:18.230 --> 00:07:23.060
vector y, then if we replace
V with that vector y, we'll

00:07:23.060 --> 00:07:25.120
still be able to generate
V because you can take a

00:07:25.120 --> 00:07:28.035
multiple of U and add
it to y and get V.

00:07:28.035 --> 00:07:31.160
And so anything that you were
able to generate with V2

00:07:31.160 --> 00:07:35.280
before, you can now generate
with our U1, multiples of

00:07:35.280 --> 00:07:37.520
that, plus multiples
of our new vector.

00:07:37.520 --> 00:07:39.570
So, we're trying to solve, we're
trying to figure out

00:07:39.570 --> 00:07:42.030
what this vector y
is right there.

00:07:42.030 --> 00:07:43.150
So how do we do it?

00:07:43.150 --> 00:07:46.790
Well, it's going to be V2
minus this vector x.

00:07:46.790 --> 00:07:47.450
Right?

00:07:47.450 --> 00:07:50.450
What is this vector
x, by definition?

00:07:50.450 --> 00:07:55.790
This vector x is, by definition,
the projection of

00:07:55.790 --> 00:08:01.550
V2 onto the subspace V1.

00:08:01.550 --> 00:08:05.550
So, the vector that we're trying
to find, the vector y,

00:08:05.550 --> 00:08:09.400
and if we find it, and that we
can replace V2 with y, the

00:08:09.400 --> 00:08:16.340
vector y, is just equal to
V2-- let me write it this

00:08:16.340 --> 00:08:23.590
way-- minus the projection
of V2 onto V1.

00:08:23.590 --> 00:08:25.550
That's what y is.

00:08:25.550 --> 00:08:29.050
Remember, if we can replace V2--
the reason why the span

00:08:29.050 --> 00:08:38.950
of U1, V2 is the same thing as
the span of U1-- let me call

00:08:38.950 --> 00:08:40.309
this y2 here.

00:08:40.309 --> 00:08:41.890
I'm going to just call that
y2, because we're probably

00:08:41.890 --> 00:08:44.230
going to have to use
y's in the future.

00:08:44.230 --> 00:08:46.610
You want y2.

00:08:46.610 --> 00:08:49.520
So the reason why this span is
the same thing as the span of

00:08:49.520 --> 00:08:56.250
U1 and y2 is because I can
generate V2 with a linear

00:08:56.250 --> 00:08:58.460
combination of U1
and y2, right?

00:08:58.460 --> 00:09:01.980
I can scale up U1 and then add
y2 and I'll get to V2.

00:09:01.980 --> 00:09:04.370
So anything that I could have
generated with V2 I can get

00:09:04.370 --> 00:09:05.930
with a linear combination
of these guys.

00:09:05.930 --> 00:09:07.590
That's why these
are equivalent.

00:09:07.590 --> 00:09:10.300
And what's neat about this now
is that these guys are

00:09:10.300 --> 00:09:12.690
orthogonal complements-- or
these guys are orthogonal

00:09:12.690 --> 00:09:14.130
relative to each other, right?

00:09:14.130 --> 00:09:17.500
By definition, y was a member of
the orthogonal complements.

00:09:17.500 --> 00:09:20.450
If you dot these two guys,
you're going to get 0.

00:09:20.450 --> 00:09:23.000
How can we actually
solve for it?

00:09:23.000 --> 00:09:27.930
Well, this is useful as
well because V1 has

00:09:27.930 --> 00:09:29.590
an orthonormal basis.

00:09:38.130 --> 00:09:41.890
And we saw, I think it was two
or three videos ago, the neat

00:09:41.890 --> 00:09:44.930
thing about orthonormal bases
is, it's very easy to

00:09:44.930 --> 00:09:47.420
determine the projection
onto those bases.

00:09:47.420 --> 00:09:52.600
It's essentially, let me write
it over here, the projection

00:09:52.600 --> 00:10:00.610
of the vector V2 onto the
subspace V1 is equal to V2--

00:10:00.610 --> 00:10:07.680
let me write it this way-- V2
dotted with the subspaces V1's

00:10:07.680 --> 00:10:14.350
first basis vector, which
is the vector U1, right?

00:10:14.350 --> 00:10:15.700
That's its first basis vector.

00:10:15.700 --> 00:10:21.070
We're dealing with an
orthonormal basis-- times U1.

00:10:21.070 --> 00:10:25.150
And then if we had more basis
vectors, we would say plus V2

00:10:25.150 --> 00:10:28.320
dotted with our next basis
vector, times that basis

00:10:28.320 --> 00:10:29.930
vector, and so on
and so forth.

00:10:29.930 --> 00:10:31.580
But V1 only has one
basis vector.

00:10:31.580 --> 00:10:34.770
It only has the basis vector
U1 right here.

00:10:34.770 --> 00:10:35.570
Right?

00:10:35.570 --> 00:10:38.270
It is only spanned by that.

00:10:38.270 --> 00:10:43.410
So, we can rewrite this thing
right here. ys, this vector

00:10:43.410 --> 00:10:47.330
right here that I'm going to
replace V2 with, is equal to

00:10:47.330 --> 00:10:55.080
V2 minus the projection of V2
onto the subspace V1, on to

00:10:55.080 --> 00:10:58.230
that line, which is just this.

00:10:58.230 --> 00:11:04.830
V2 dot U1 times the vector U1.

00:11:04.830 --> 00:11:06.910
And just like that, we've
solved for y.

00:11:06.910 --> 00:11:11.690
So we have a basis for V2, where
this guy and this guy

00:11:11.690 --> 00:11:12.710
are orthogonal.

00:11:12.710 --> 00:11:13.870
This guy is the unit vector.

00:11:13.870 --> 00:11:15.120
He's been normalized,
but this guy hasn't

00:11:15.120 --> 00:11:16.660
been normalized yet.

00:11:16.660 --> 00:11:24.410
So to normalize it, let's just
define some other vector as U2

00:11:24.410 --> 00:11:25.100
and do the same thing.

00:11:25.100 --> 00:11:26.060
Let's just normalize it.

00:11:26.060 --> 00:11:30.650
So U2 is equal to y2, divided
by the length of y2.

00:11:35.050 --> 00:11:38.000
Now, we can say that the
subspace V2 is equal to the

00:11:38.000 --> 00:11:45.330
span of U1 and, instead of y2,
I'll put U2 there, because I

00:11:45.330 --> 00:11:49.990
can generate y2 by just
scaling up U2.

00:11:49.990 --> 00:11:53.130
What's neat about this now is,
I have two unit vectors, or

00:11:53.130 --> 00:11:55.380
two normalized vectors, and
they're orthogonal with

00:11:55.380 --> 00:11:58.460
respect to each and they span
what V1 and V2 spanned

00:11:58.460 --> 00:11:59.590
originally.

00:11:59.590 --> 00:12:00.280
Now,

00:12:00.280 --> 00:12:01.930
we need to keep going.

00:12:01.930 --> 00:12:06.100
What about if we want to go
to V3 here, what do we do?

00:12:06.100 --> 00:12:09.730
Well, we do the same idea.

00:12:09.730 --> 00:12:13.490
So let's define the
subspace V3.

00:12:13.490 --> 00:12:16.430
This is going to be a
three-dimensional subspace.

00:12:16.430 --> 00:12:18.340
Once we get beyond V3, it
becomes a little hard to

00:12:18.340 --> 00:12:19.720
visualize, but I think you're
going to see the

00:12:19.720 --> 00:12:21.240
pattern after this step.

00:12:21.240 --> 00:12:29.920
If we define V3 as equal to the
span of these guys, U1,

00:12:29.920 --> 00:12:34.420
U2, and then, in our
original basis, V3.

00:12:34.420 --> 00:12:37.980
I didn't write it there but
there's a V3 there.

00:12:37.980 --> 00:12:40.150
So in our original
non-orthonormal

00:12:40.150 --> 00:12:42.980
basis, we have V3.

00:12:42.980 --> 00:12:45.110
What is this going
to look like?

00:12:45.110 --> 00:12:47.870
Or how can we turn this into
an orthonormal basis?

00:12:47.870 --> 00:12:53.060
So, if you visualize all of
these, so the span of U1 and

00:12:53.060 --> 00:12:57.460
U2, our subspace V2 is just
going to be a plane.

00:12:57.460 --> 00:12:59.720
It's going to be
a plane in R3.

00:12:59.720 --> 00:13:03.950
It's going to look like
that right there.

00:13:03.950 --> 00:13:07.430
And so, our new span is going
to be everything in that

00:13:07.430 --> 00:13:10.370
plane, all linear combinations
of things in that plane, plus

00:13:10.370 --> 00:13:14.010
linear combinations of our
vector V3, which is linearly

00:13:14.010 --> 00:13:15.930
independent from these guys
because it was linearly

00:13:15.930 --> 00:13:19.220
independent from the stuff we
used to construct these guys.

00:13:19.220 --> 00:13:21.605
So V3 is going to pop
out of this plane.

00:13:21.605 --> 00:13:23.170
It can't be represented
by the linear

00:13:23.170 --> 00:13:25.610
combination of those guys.

00:13:25.610 --> 00:13:29.650
So let's say V3 pops out of
the plane, like that.

00:13:29.650 --> 00:13:33.730
Now, we want another vector that
can represent everything

00:13:33.730 --> 00:13:37.760
of this span, but it is
orthogonal to these dudes

00:13:37.760 --> 00:13:38.920
right here.

00:13:38.920 --> 00:13:40.190
Or another way to think
about it, it's

00:13:40.190 --> 00:13:41.320
orthogonal to the plane.

00:13:41.320 --> 00:13:44.990
So we want another vector that
is orthogonal to the plane.

00:13:44.990 --> 00:13:51.460
So let's call that vector y3,
y sub 3, not y to the third

00:13:51.460 --> 00:13:53.340
power, y sub 3.

00:13:53.340 --> 00:13:59.420
And if we figure out our y sub
three, we can replace V3 with

00:13:59.420 --> 00:14:01.570
it because V3 can be represented
as a linear

00:14:01.570 --> 00:14:03.250
combination of U1 and U2.

00:14:03.250 --> 00:14:03.890
Right?

00:14:03.890 --> 00:14:06.850
That's going to be a linear
combination of U1, U2 is going

00:14:06.850 --> 00:14:10.810
to be some vector in
this plane plus y3.

00:14:10.810 --> 00:14:13.460
I can represent this guy with
this green vector plus this

00:14:13.460 --> 00:14:13.910
vector right here.

00:14:13.910 --> 00:14:17.370
So if we replace him with y3, we
can still get to V3 and we

00:14:17.370 --> 00:14:19.600
can still get to all the linear
combinations that V3

00:14:19.600 --> 00:14:21.020
can help construct.

00:14:21.020 --> 00:14:23.040
So what is this y3?

00:14:23.040 --> 00:14:26.160
Well, by the same exact logic,
this green vector right here

00:14:26.160 --> 00:14:37.110
is the projection of my vector
V3 onto the subspace V2 and

00:14:37.110 --> 00:14:44.300
your vector y3 is just equal
to the vector V3 minus the

00:14:44.300 --> 00:14:51.360
projection of V3 onto V2.

00:14:51.360 --> 00:14:53.370
So, what is that going to be?

00:14:53.370 --> 00:14:56.950
Well, the projection-- I'll
write it here-- the projection

00:14:56.950 --> 00:15:03.560
onto the subspace V2 of V3, of
the vector V3 is going to be

00:15:03.560 --> 00:15:05.380
equal to-- we're going
to use the same exact

00:15:05.380 --> 00:15:06.310
logic we did before.

00:15:06.310 --> 00:15:08.140
We saw this two or
three videos ago.

00:15:08.140 --> 00:15:13.170
Because V2 is defined with an
orthonormal basis, we can say

00:15:13.170 --> 00:15:19.260
that the projection of V3 onto
that subspace is V3, dot our

00:15:19.260 --> 00:15:25.460
first basis vector, dot U1,
times our first basis vector,

00:15:25.460 --> 00:15:32.550
plus V3 dot our second basis
vector, our second orthonormal

00:15:32.550 --> 00:15:36.090
basis vector, times our second
orthonormal basis vector.

00:15:36.090 --> 00:15:36.770
It's that easy.

00:15:36.770 --> 00:15:38.440
That's one of the neat things
about having an

00:15:38.440 --> 00:15:39.680
orthonormal basis was.

00:15:39.680 --> 00:15:41.200
This is an orthonormal basis.

00:15:41.200 --> 00:15:43.870
So we can define the projection
in this way.

00:15:43.870 --> 00:15:47.570
So y is just going to be V3
minus this business right

00:15:47.570 --> 00:15:49.460
here, and I'll just write out.

00:15:49.460 --> 00:15:59.910
So y3 is going to be V3 minus
the projection of V3 onto V2.

00:15:59.910 --> 00:16:04.690
So minus that business
right there.

00:16:04.690 --> 00:16:08.700
Let me copy it, and then
let me paste it.

00:16:08.700 --> 00:16:11.770
And then you get y3.

00:16:11.770 --> 00:16:13.105
And y3 is nice.

00:16:13.105 --> 00:16:16.310
So if we replace this with y3,
which we can because we can

00:16:16.310 --> 00:16:18.970
now get V3 as a linear
combination of

00:16:18.970 --> 00:16:21.520
these guys and y3.

00:16:21.520 --> 00:16:23.870
That's nice because all of these
guys are orthogonal with

00:16:23.870 --> 00:16:28.500
respect to each other, but y3
does not have length 1 yet.

00:16:28.500 --> 00:16:29.630
It hasn't been normalized.

00:16:29.630 --> 00:16:32.950
So we can just replace y3 with
another unit vector.

00:16:32.950 --> 00:16:38.950
We'll just say, U3 is equal to
y3 divided by the length of

00:16:38.950 --> 00:16:41.450
y3, whatever that might be.

00:16:41.450 --> 00:16:43.880
But if we have what y3 is, we
can figure out its length and

00:16:43.880 --> 00:16:45.080
divide by that.

00:16:45.080 --> 00:16:50.550
And then if you put-- so this
is equal to the span of U1,

00:16:50.550 --> 00:16:54.700
U2, and U3.

00:16:54.700 --> 00:16:59.300
So U3 might be a scaled-down
version of y3.

00:16:59.300 --> 00:17:00.860
We're talking about linear
combinations when we talk

00:17:00.860 --> 00:17:03.070
about span, so you can scale
them up and add it to the

00:17:03.070 --> 00:17:03.480
projection.

00:17:03.480 --> 00:17:04.589
You're still going to get V3.

00:17:04.589 --> 00:17:07.190
And now, these are all
normalized vectors.

00:17:07.190 --> 00:17:11.720
And so you would not have an
orthonormal basis for the

00:17:11.720 --> 00:17:13.119
subspace V3.

00:17:13.119 --> 00:17:14.589
I think you get the
pattern now.

00:17:14.589 --> 00:17:15.935
You can keep doing this.

00:17:18.520 --> 00:17:23.950
You can define V4, when I
defined the problem I just

00:17:23.950 --> 00:17:27.030
said that we're dealing with
a k dimensional subspace.

00:17:27.030 --> 00:17:27.930
We go up to Vk.

00:17:27.930 --> 00:17:29.580
But you just keep going until
you go up to your k.

00:17:29.580 --> 00:17:31.570
If k was 3, we'd be done.

00:17:31.570 --> 00:17:38.650
If k is 4, you say you define a
subspace V4 is equal to the

00:17:38.650 --> 00:17:47.290
span of U1, U2, U3, and then
throw our non-orthonormal

00:17:47.290 --> 00:17:49.700
vector, V4, into the mix.

00:17:49.700 --> 00:17:59.180
And so you can replace V4 with
y4 is equal V4 minus the

00:17:59.180 --> 00:18:00.080
projection.

00:18:00.080 --> 00:18:01.370
It becomes a little
hard to visualize.

00:18:01.370 --> 00:18:03.720
You're not projecting on to
a three-dimensional space.

00:18:03.720 --> 00:18:10.280
The projection of V4 onto
our subspace V3.

00:18:10.280 --> 00:18:12.120
It's completely analogous
to this.

00:18:12.120 --> 00:18:14.370
It's just V3 is now a
three-dimensional

00:18:14.370 --> 00:18:15.190
space, not a plane.

00:18:15.190 --> 00:18:15.690
You're [? finding ?]

00:18:15.690 --> 00:18:16.700
this projection onto this.

00:18:16.700 --> 00:18:21.300
This is definitely going to be
orthogonal to everything else,

00:18:21.300 --> 00:18:23.090
to our subspace V3.

00:18:23.090 --> 00:18:31.560
And you can construct V4 with y4
because V4, by definition,

00:18:31.560 --> 00:18:36.120
is equal to-- we can just
rearrange this-- y4 plus the

00:18:36.120 --> 00:18:40.480
projection of V4 onto V3.

00:18:40.480 --> 00:18:44.275
So you can construct V4 with y4
and some linear combination

00:18:44.275 --> 00:18:45.140
of these guys.

00:18:45.140 --> 00:18:47.860
So I can replace this guy
with y4 and then I

00:18:47.860 --> 00:18:48.920
would normalize y4.

00:18:48.920 --> 00:18:51.145
I would divide it by its
length and I'd get U4.

00:18:51.145 --> 00:18:54.660
And I just keep doing that
until I get up to k.

00:18:54.660 --> 00:18:58.290
If I do V5, I do the process
over and over and over again.

00:18:58.290 --> 00:19:05.060
And this process of creating an
orthonormal basis is called

00:19:05.060 --> 00:19:12.275
the Gram-Schmidt Process.

00:19:12.275 --> 00:19:14.760
And it might seem a little
abstract, the way I did it

00:19:14.760 --> 00:19:17.480
here, but in the next video
I'm actually going to find

00:19:17.480 --> 00:19:19.820
orthonormal bases
for subspaces.

00:19:19.820 --> 00:19:21.640
You'll see it's not too bad
when you have to deal with

00:19:21.640 --> 00:19:23.110
real numbers.


Orthogonal matrices preserve angles and lengths

WEBVTT
Kind: captions
Language: en

00:00:00.580 --> 00:00:02.960
In the last couple of videos,
we've seen that if we have

00:00:02.960 --> 00:00:06.460
some matrix C that is n by n.

00:00:06.460 --> 00:00:12.430
It's a square matrix, and is
columns, column form and

00:00:12.430 --> 00:00:13.790
orthonormal set.

00:00:18.750 --> 00:00:22.110
Which just means that the
columns each have been

00:00:22.110 --> 00:00:22.670
normalized.

00:00:22.670 --> 00:00:24.610
So they each have length
of 1 if you view

00:00:24.610 --> 00:00:25.940
them as column vectors.

00:00:25.940 --> 00:00:28.520
And they're all mutually
orthogonal to each other.

00:00:28.520 --> 00:00:30.930
So if you dot it with
yourself you get 1.

00:00:30.930 --> 00:00:33.480
If you dot it with any of the
other columns, you get 0.

00:00:33.480 --> 00:00:34.670
We've seen this multiple
times.

00:00:34.670 --> 00:00:36.520
It's orthogonal to
everything else.

00:00:36.520 --> 00:00:39.400
If you have a matrix like this--
and I actually forgot

00:00:39.400 --> 00:00:41.270
to tell you the name of
this-- this is called

00:00:41.270 --> 00:00:42.520
an orthogonal matrix.

00:00:47.130 --> 00:00:50.950
We've already seen that the
transpose of this matrix is

00:00:50.950 --> 00:00:53.530
the same thing as the inverse
of this matrix.

00:00:53.530 --> 00:00:57.320
Which makes it super, duper,
duper useful to deal with.

00:00:57.320 --> 00:01:00.600
The transpose of this matrix
is equal to the inverse.

00:01:00.600 --> 00:01:03.650
Now, this statement leads to
some other interesting things

00:01:03.650 --> 00:01:04.160
about this.

00:01:04.160 --> 00:01:06.400
So, so far we've been dealing
this mainly with

00:01:06.400 --> 00:01:07.510
the change of basis.

00:01:07.510 --> 00:01:09.460
I can kind of draw the
diagram that you're

00:01:09.460 --> 00:01:12.580
probably tired of by now.

00:01:12.580 --> 00:01:14.290
Let's say that's the
standard basis.

00:01:14.290 --> 00:01:17.560
Let's say that I have
x in coordinates

00:01:17.560 --> 00:01:18.760
with another basis.

00:01:18.760 --> 00:01:22.280
We've seen I can multiply
this guy times c.

00:01:22.280 --> 00:01:25.720
To get that up there I could
multiply that guy by c inverse

00:01:25.720 --> 00:01:27.090
to get this guy right here.

00:01:27.090 --> 00:01:30.600
And, in that world, we viewed
c as just a change of basis.

00:01:30.600 --> 00:01:33.260
Were representing the same
matrix-- we're representing

00:01:33.260 --> 00:01:34.280
the same vector.

00:01:34.280 --> 00:01:36.100
We're just changing the
coordinates of how we

00:01:36.100 --> 00:01:36.970
represent it.

00:01:36.970 --> 00:01:40.220
But we also know that any matrix
product, any matrix

00:01:40.220 --> 00:01:43.080
vector product, is also a
linear transformation.

00:01:43.080 --> 00:01:45.820
So, this change of basis
is really just a linear

00:01:45.820 --> 00:01:47.340
transformation.

00:01:47.340 --> 00:01:48.890
What I want to show you in this
video, and you could view

00:01:48.890 --> 00:01:52.930
it either as a change of basis
or as a linear transformation,

00:01:52.930 --> 00:02:00.230
is that when you multiply this
orthogonal matrix times some

00:02:00.230 --> 00:02:03.170
vector, it preserves--
let me write this

00:02:03.170 --> 00:02:13.880
down-- lengths and angles.

00:02:13.880 --> 00:02:15.880
So let's have a little
touchy-feely discussion of

00:02:15.880 --> 00:02:17.390
what that means.

00:02:17.390 --> 00:02:19.390
Let's view it as a
transformation.

00:02:19.390 --> 00:02:23.070
Let's say I have some set
of vectors in my domain.

00:02:23.070 --> 00:02:25.320
Let's say they look like this.

00:02:25.320 --> 00:02:28.330
Let's say that it
looks like this.

00:02:28.330 --> 00:02:31.210
Well, let me do it like-- I'll
draw that one like that guy,

00:02:31.210 --> 00:02:33.365
and this guy like that.

00:02:33.365 --> 00:02:36.150
And there's some angle
between them.

00:02:36.150 --> 00:02:38.610
Angles are easy to visualize
in r2, r3.

00:02:38.610 --> 00:02:40.780
Maybe a little harder once we
get the higher dimensions.

00:02:40.780 --> 00:02:42.820
But that's the angle
between them.

00:02:42.820 --> 00:02:46.980
Now, if we're saying that we're
preserving the angles

00:02:46.980 --> 00:02:49.830
and the lengths, that means if
I were to multiply these

00:02:49.830 --> 00:02:53.440
vectors times c then we could
view it as a transformation.

00:02:53.440 --> 00:02:56.650
Maybe I rotate them or I--
well, you can't really.

00:02:56.650 --> 00:02:58.530
Maybe I rotate them or do
something like that.

00:02:58.530 --> 00:03:02.490
So maybe that pink vector
will now look like this.

00:03:02.490 --> 00:03:03.700
But it's going to have
the same length.

00:03:03.700 --> 00:03:05.990
This length is going to be the
same thing as that length.

00:03:05.990 --> 00:03:08.575
And even more, when I said it
preserves lengths and angles,

00:03:08.575 --> 00:03:11.840
this yellow vector's going to
look something like this.

00:03:11.840 --> 00:03:14.630
Where the angle is going
to be the same.

00:03:14.630 --> 00:03:16.980
Where this data is going
to be that data.

00:03:16.980 --> 00:03:19.200
That's what I mean by
preserves angles.

00:03:19.200 --> 00:03:21.530
If we didn't have this case,
we could imagine a

00:03:21.530 --> 00:03:23.460
transformation that doesn't
preserve angles.

00:03:23.460 --> 00:03:24.690
Let me draw one that doesn't.

00:03:24.690 --> 00:03:28.250
If this got transformed to, I
don't know, let's say this guy

00:03:28.250 --> 00:03:33.460
got a lot longer, and let's say
this guy also got longer,

00:03:33.460 --> 00:03:35.330
and I want to show
that the angle

00:03:35.330 --> 00:03:36.235
also doesn't get preserved.

00:03:36.235 --> 00:03:37.400
Not only did it get
longer, but it got

00:03:37.400 --> 00:03:38.370
distorted a little bit.

00:03:38.370 --> 00:03:41.380
So, the angle also changed.

00:03:41.380 --> 00:03:43.850
This transformation
right there is

00:03:43.850 --> 00:03:50.760
not preserving angles.

00:03:50.760 --> 00:03:54.850
So when you have a change
of basis matrix that's

00:03:54.850 --> 00:03:58.010
orthogonal, when you have a
transformation matrix that's

00:03:58.010 --> 00:04:02.220
orthogonal, all it's essentially
doing to your to

00:04:02.220 --> 00:04:05.820
your vectors, is it kind of a
rotates them around, but it's

00:04:05.820 --> 00:04:08.110
not going to really
distort them.

00:04:08.110 --> 00:04:10.960
So I'll write that in quotes
because that's not a

00:04:10.960 --> 00:04:12.430
mathematically rigorous term.

00:04:12.430 --> 00:04:18.690
So, no distortion of vectors.

00:04:18.690 --> 00:04:20.570
So, I've kind of showed
you the intuition

00:04:20.570 --> 00:04:21.220
of what that means.

00:04:21.220 --> 00:04:22.745
Let's actually prove
it to ourselves

00:04:22.745 --> 00:04:24.980
that this is the case.

00:04:24.980 --> 00:04:30.180
So, I'm saying that if this
pink vector here is x, and

00:04:30.180 --> 00:04:34.630
that this pink vector here is
c times x, I'm claiming that

00:04:34.630 --> 00:04:39.400
the length of x is equal to
the length of c times x.

00:04:39.400 --> 00:04:41.373
Let's see if that's
actually the case.

00:04:47.180 --> 00:04:56.830
The length of cx squared is the
same thing as cx dot cx.

00:04:56.830 --> 00:04:59.360
And here it's always useful
for me to kind of remind

00:04:59.360 --> 00:05:01.940
myself that if I take
two vectors-- let

00:05:01.940 --> 00:05:02.810
me do it over here.

00:05:02.810 --> 00:05:06.810
Let's say I have y dot y.

00:05:06.810 --> 00:05:11.593
This is the same thing as y
transpose, if you view them as

00:05:11.593 --> 00:05:15.300
matrices, y transpose times y.

00:05:15.300 --> 00:05:24.180
y transpose y is just y1, y2,
all the way to yn times y1,

00:05:24.180 --> 00:05:26.340
y2, all the way to yn.

00:05:26.340 --> 00:05:31.610
And if you were to do this 1
by n times n by 1 matrix

00:05:31.610 --> 00:05:35.340
product, you're going to get a 1
by 1 matrix or just a number

00:05:35.340 --> 00:05:38.770
that's going to be y1 times y1
plus y2 times y2 all the way

00:05:38.770 --> 00:05:39.910
to yn times yn.

00:05:39.910 --> 00:05:44.530
So, this is the same
thing as y dot y.

00:05:44.530 --> 00:05:46.980
I think I did this about ten or
twenty videos ago, but it's

00:05:46.980 --> 00:05:48.630
always a good refresher.

00:05:48.630 --> 00:05:50.170
So let's use this property
right here.

00:05:50.170 --> 00:05:52.110
So these two dotted
with each other.

00:05:52.110 --> 00:05:55.300
This is the same thing is taking
one of their transpose

00:05:55.300 --> 00:05:56.390
times the other one.

00:05:56.390 --> 00:06:00.420
So turn this from a vector,
vector dot product to a

00:06:00.420 --> 00:06:02.580
matrix, matrix product.

00:06:02.580 --> 00:06:07.930
So this is the same thing
as CX transpose, CX.

00:06:07.930 --> 00:06:13.480
so you can view this as a 1 by n
matrix now, times the 1 by 1

00:06:13.480 --> 00:06:16.110
matrix which is just the
column vector cx.

00:06:16.110 --> 00:06:17.760
These are the same thing.

00:06:17.760 --> 00:06:24.330
Now, we also know that A times
B transpose is the same thing

00:06:24.330 --> 00:06:26.430
is B transpose, A transpose.

00:06:26.430 --> 00:06:28.240
We saw that a long time ago.

00:06:28.240 --> 00:06:30.710
So this thing right here is
going to be equal to X

00:06:30.710 --> 00:06:32.890
transpose, C transpose.

00:06:32.890 --> 00:06:35.580
Just switch the order and take
the transpose of each.

00:06:35.580 --> 00:06:37.710
X transpose times C transpose.

00:06:37.710 --> 00:06:43.170
And then you have
that times CX.

00:06:43.170 --> 00:06:46.180
And now we know that C transpose
is the same thing is

00:06:46.180 --> 00:06:47.430
as C inverse.

00:06:49.600 --> 00:06:52.690
This is where we need the
orthogonality of the matrix C.

00:06:52.690 --> 00:06:55.130
This is where we need it to be
a square matrix where all of

00:06:55.130 --> 00:06:57.540
its columns are mutually
orthogonal

00:06:57.540 --> 00:06:58.630
and they're all normal.

00:06:58.630 --> 00:07:00.630
And so this thing is just
going to become

00:07:00.630 --> 00:07:01.880
the identity matrix.

00:07:05.370 --> 00:07:07.240
I can write the identity matrix
there, but that's just

00:07:07.240 --> 00:07:08.010
going to disappear.

00:07:08.010 --> 00:07:12.680
So this is going to be equal
to X transpose X.

00:07:12.680 --> 00:07:18.190
X transpose is the same thing
as X dot X which is the same

00:07:18.190 --> 00:07:21.840
thing as the length
of X squared.

00:07:21.840 --> 00:07:24.130
So the length of CX squared is
the same thing as the length

00:07:24.130 --> 00:07:25.140
of X squared.

00:07:25.140 --> 00:07:27.770
So, that tells us that the
length of X, or the length of

00:07:27.770 --> 00:07:30.700
CX, is the length of x because
both of these are going to be

00:07:30.700 --> 00:07:32.280
positive quantities.

00:07:32.280 --> 00:07:36.750
So I've shown you that
orthogonal matrices definitely

00:07:36.750 --> 00:07:37.760
preserve length.

00:07:37.760 --> 00:07:39.990
Let's see if they
preserve angles.

00:07:39.990 --> 00:07:43.750
So we actually have
to define angles.

00:07:43.750 --> 00:07:46.260
Throughout our mathematical
careers, we understood what

00:07:46.260 --> 00:07:48.520
angles mean in kind
of r2 or r3.

00:07:48.520 --> 00:07:50.440
But in linear algebra, we
like to be general.

00:07:50.440 --> 00:07:53.710
And we defined an angle
using the dot product.

00:07:53.710 --> 00:07:57.650
We use the law of cosines and we
took an analogy to kind of

00:07:57.650 --> 00:07:58.640
triangle in r2.

00:07:58.640 --> 00:08:07.760
But we defined an angle or we
said the dot product V dot W

00:08:07.760 --> 00:08:14.680
is equal to the lengths, the
products of the lengths of

00:08:14.680 --> 00:08:17.310
those two vectors times
the cosine of the

00:08:17.310 --> 00:08:18.720
angle between them.

00:08:18.720 --> 00:08:22.350
Or you could say that the cosine
of the angle between

00:08:22.350 --> 00:08:26.780
two vectors, we defined as the
dot product of those two

00:08:26.780 --> 00:08:29.710
vectors divided by the lengths
of those two vectors.

00:08:29.710 --> 00:08:33.289
This was the definition so that
we can extend the idea of

00:08:33.289 --> 00:08:37.240
an angle to an arbitrarily
high dimension to r

00:08:37.240 --> 00:08:39.610
google if we had to.

00:08:39.610 --> 00:08:41.860
So let's see if it preserves.

00:08:41.860 --> 00:08:45.860
Let's see what the angle is if
we multiply these guys by C.

00:08:45.860 --> 00:08:48.010
So, if we wanted-- let's
say our new angle.

00:08:48.010 --> 00:08:51.370
So, cosine of angle C.

00:08:51.370 --> 00:08:54.740
Once we perform our
transformation.

00:08:54.740 --> 00:08:56.690
We're going to perform the
transformation on all of these

00:08:56.690 --> 00:08:57.230
characters.

00:08:57.230 --> 00:09:05.270
It's going to be CV dot CW over
the lengths of CV times

00:09:05.270 --> 00:09:08.370
the lengths of CW.

00:09:08.370 --> 00:09:10.490
Now we already know that
lengths are preserved.

00:09:10.490 --> 00:09:12.930
We already know that the length
of CW and CV are just

00:09:12.930 --> 00:09:13.920
going to be W and V.

00:09:13.920 --> 00:09:14.780
We just proved that.

00:09:14.780 --> 00:09:15.600
Let me write that.

00:09:15.600 --> 00:09:22.950
So the cosine of theta C is
equal to CV dot CW over the

00:09:22.950 --> 00:09:26.450
lengths of V times W.

00:09:26.450 --> 00:09:28.670
Because we've already shown
that it preserves lengths.

00:09:28.670 --> 00:09:31.040
We'll see what this
top part equals.

00:09:31.040 --> 00:09:34.030
So we can just use the
general property.

00:09:34.030 --> 00:09:37.330
The dot product is equal to the
transpose of one guy as

00:09:37.330 --> 00:09:39.010
kind of a matrix times
the second guy.

00:09:39.010 --> 00:09:47.370
So this is equal to CW
transpose times CV.

00:09:47.370 --> 00:09:51.220
And all of that over
these lengths.

00:09:51.220 --> 00:09:53.060
Like the W.

00:09:53.060 --> 00:09:56.150
And this is going to be equal
to-- I'm going to write it

00:09:56.150 --> 00:09:57.760
down here to have some space.

00:09:57.760 --> 00:10:00.040
We can switch these guys and
take their transpose.

00:10:00.040 --> 00:10:06.720
So, it's W transpose times
C transpose times CV.

00:10:06.720 --> 00:10:09.440
All of that over their lengths,
the product of their

00:10:09.440 --> 00:10:11.990
lengths V and W.

00:10:11.990 --> 00:10:13.860
And this is the identity
matrix.

00:10:13.860 --> 00:10:15.930
That's the identity matrix, and
this is going to be equal

00:10:15.930 --> 00:10:20.880
to W transpose times V over the
products of their lengths.

00:10:24.300 --> 00:10:28.560
And this is the same
thing as V dot W.

00:10:28.560 --> 00:10:34.950
This is V dot W over
their lengths.

00:10:34.950 --> 00:10:38.750
Which is cosine of theta.

00:10:38.750 --> 00:10:42.170
So, you notice, by our
definition of an angle as the

00:10:42.170 --> 00:10:45.040
dot product divided by the
vector lengths, when you

00:10:45.040 --> 00:10:48.190
perform a transformation or you
can imagine a change of

00:10:48.190 --> 00:10:54.750
basis either way, with an
orthogonal matrix C the angle

00:10:54.750 --> 00:10:57.940
between the transformed vectors
does not change.

00:10:57.940 --> 00:11:00.640
It is the same as the angle
between the vectors before

00:11:00.640 --> 00:11:01.980
they were transformed.

00:11:01.980 --> 00:11:03.340
Which is a really neat
thing to know.

00:11:03.340 --> 00:11:05.640
The change of bases or
transformations with

00:11:05.640 --> 00:11:08.970
orthogonal matrices don't
distort the vectors.

00:11:08.970 --> 00:11:11.300
They might just kind of rotate
them around or shift them a

00:11:11.300 --> 00:11:12.540
little bit, but it doesn't
change the

00:11:12.540 --> 00:11:14.700
angles between them.