Showing that inverses are linear

WEBVTT
Kind: captions
Language: en

00:00:00.780 --> 00:00:03.190
I've got a transformation T.

00:00:03.190 --> 00:00:06.670
When you apply the
transformation T to some x in

00:00:06.670 --> 00:00:10.890
your domain, it is equivalent to
multiplying that x in your

00:00:10.890 --> 00:00:12.220
domain, or that vector.

00:00:12.220 --> 00:00:14.380
by the matrix A.

00:00:14.380 --> 00:00:17.950
And let's say we know the linear
transformation T can

00:00:17.950 --> 00:00:20.940
be-- that's transformation
matrix when you put it in

00:00:20.940 --> 00:00:29.220
reduced row echelon form-- it is
equal to an n by n identity

00:00:29.220 --> 00:00:32.560
matrix, or the n by
n identity matrix.

00:00:32.560 --> 00:00:35.230
Well this alone tells
a lot of things.

00:00:35.230 --> 00:00:37.800
First of all, if when you put
this in reduced row echelon

00:00:37.800 --> 00:00:40.000
form, you get a square
identity matrix.

00:00:40.000 --> 00:00:41.650
That tells us that the
original matrix

00:00:41.650 --> 00:00:43.230
had to be n by n.

00:00:43.230 --> 00:00:49.600
And it also tells us that T is
a mapping from Rn to Rn.

00:00:49.600 --> 00:00:53.530
And we saw in the last video,
all of these are conditions,

00:00:53.530 --> 00:00:59.010
especially this one right here,
for T to be invertible.

00:00:59.010 --> 00:01:00.790
So If we know that this is
true-- T is a linear

00:01:00.790 --> 00:01:03.390
transformation, it's a reduced
row echelon form of the

00:01:03.390 --> 00:01:06.460
transformation matrix of the
identity matrix right there--

00:01:06.460 --> 00:01:09.620
we know that T is invertible.

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Let's remind ourselves what it
even means to be invertible.

00:01:15.910 --> 00:01:23.910
To be invertible means that
there exists some-- we used

00:01:23.910 --> 00:01:25.720
the word function before, but
now we're talking about

00:01:25.720 --> 00:01:27.580
transformations, they're
really the same thing.

00:01:27.580 --> 00:01:35.760
So let's say there exists some
transformation-- let's call it

00:01:35.760 --> 00:01:46.050
T-inverse like that, or T to
the minus 1-- such that the

00:01:46.050 --> 00:01:51.750
composition of T-inverse with
T is equal to the identity

00:01:51.750 --> 00:02:00.350
transformation on your domain,
and the composition of T with

00:02:00.350 --> 00:02:05.010
T-inverse is equal to the
identity transformation on

00:02:05.010 --> 00:02:06.390
your codomain.

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Just like that.

00:02:09.620 --> 00:02:12.720
And just to remind you what this
looks like, let's draw

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our domains and codomains.

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Our domain is Rn, and our
codomain is also Rn.

00:02:23.250 --> 00:02:28.520
So if you take some vector
in your domain, apply the

00:02:28.520 --> 00:02:30.420
transformation T-- you're
going to go into your

00:02:30.420 --> 00:02:32.310
codomain, so that is T.

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And then if you apply the
T-inverse after that, you're

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going to go back to
that original x.

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So this says, look,
you apply T and

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then you apply T-inverse.

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You're just going to get back
to where you started.

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It's equivalent to the identity
transformation.

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Just like that.

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This is saying if you start
here in your codomain, you

00:02:55.990 --> 00:02:59.080
apply this inverse
transformation first, then you

00:02:59.080 --> 00:03:01.670
apply your transformation,
you're going to go back to the

00:03:01.670 --> 00:03:03.280
same point in your codomain.

00:03:03.280 --> 00:03:06.350
So it's equivalent to the
identity transformation in

00:03:06.350 --> 00:03:07.640
your codomain.

00:03:07.640 --> 00:03:09.710
It just happens to be in this
case that the domain and the

00:03:09.710 --> 00:03:14.470
codomain are the same set, Rn.

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Now we know what a
transformation-- what it means

00:03:17.740 --> 00:03:20.420
to be invertible. we know what
the conditions are for

00:03:20.420 --> 00:03:21.040
invertibility.

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So this begs the
next question.

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We know this guy is a linear
transformation, in fact that's

00:03:26.570 --> 00:03:27.760
one of the conditions
to be able to

00:03:27.760 --> 00:03:28.910
represent it as a matrix.

00:03:28.910 --> 00:03:31.000
Or any transformation that can
be represented as a matrix

00:03:31.000 --> 00:03:33.550
vector product is a linear
transformation.

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So this guy's a linear
transformation.

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But the question is, is
T-inverse a linear

00:03:42.920 --> 00:03:44.170
transformation?

00:03:48.660 --> 00:03:51.750
Let's review what the two
conditions are that we need to

00:03:51.750 --> 00:03:54.610
have to be a linear
transformation.

00:03:54.610 --> 00:03:56.400
So we know T is a linear
transformation.

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So we know that if you apply
the transformation T to two

00:04:00.150 --> 00:04:03.530
vectors-- let's say x and y--
if we apply to the sum of

00:04:03.530 --> 00:04:07.390
those two vectors, it is equal
to the transformation of the

00:04:07.390 --> 00:04:09.650
first vector plus the
transformation

00:04:09.650 --> 00:04:11.030
of the second vector.

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That's one of the conditions,
or one thing that we know is

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true for all linear
transformations.

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And the second thing we know
is true for all linear

00:04:18.709 --> 00:04:22.560
transformations is, if we take
the transformation of some

00:04:22.560 --> 00:04:27.500
scaled version of a vector in
our domain, it is equal to the

00:04:27.500 --> 00:04:30.680
scaling factor times
the transformation

00:04:30.680 --> 00:04:32.270
of the vector itself.

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These are both conditions for
linear transformations.

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So let's see if we can prove
that both of these conditions

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hold for T-inverse, or
this guy right here.

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So to do this, let's do this
little exercise right here.

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Let's take the composition of
T with T-inverse of two

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vectors, a plus b.

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Remember, T-inverse is a mapping
from your codomain to

00:05:04.585 --> 00:05:05.980
your domain, although they're
both going to

00:05:05.980 --> 00:05:07.580
be Rn in this case.

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But T-inverse maps from
this set to that set.

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Let's write it up here.

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T-inverse is a mapping from your
codomain to your domain.

00:05:15.470 --> 00:05:18.100
Although it looks identical,
just like that.

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OK, so what is this going
to be equal to?

00:05:20.560 --> 00:05:24.820
Well we just said, by definition
of your inverse

00:05:24.820 --> 00:05:27.040
transformation, this is going
to be equal to the identity

00:05:27.040 --> 00:05:29.360
transformation on
your codomain.

00:05:29.360 --> 00:05:31.860
So assuming these guys are
members of your codomain, in

00:05:31.860 --> 00:05:41.180
this case Rn, this is just going
to be equal to a plus b.

00:05:41.180 --> 00:05:43.640
This thing, the composition
of T with its inverse, by

00:05:43.640 --> 00:05:46.090
definition is just the identity
transformation on

00:05:46.090 --> 00:05:47.270
your codomain.

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So this is just whatever
I put in here.

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If I put in an x here,
this would be an x.

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If I put in an apple here,
this would be an apple.

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It's going to be the identity
transformation.

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Now what is this equal to?

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Well I could use the same
argument to say that this

00:06:02.940 --> 00:06:08.290
right here is equal to the
identity transformation

00:06:08.290 --> 00:06:10.630
applied to a.

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And I'm not writing the identity
transformation, I'm

00:06:12.750 --> 00:06:13.330
writing this.

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But we know that this is
equivalent to the identity

00:06:15.580 --> 00:06:16.320
transformation.

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So we could say that is
equivalent to the composition

00:06:19.100 --> 00:06:22.640
of T with the inverse applied
to a, and we could say that

00:06:22.640 --> 00:06:26.900
this is the equivalent to the
identity transformation, which

00:06:26.900 --> 00:06:30.510
we know is the same thing as T,
the composition of t with

00:06:30.510 --> 00:06:33.210
T-inverse applied to b.

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So we can rewrite this thing
right here as being equal to

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the sum of these two things.

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In fact we don't even
have to rewrite it.

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We can just write it's equal
to-- this transformation is

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equal to this.

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And maybe an easier way for you
to process it is, we could

00:06:54.520 --> 00:07:02.340
write this as T of the T-inverse
of a plus b, is

00:07:02.340 --> 00:07:10.700
equal to T of the T-inverse
of a plus T of the

00:07:10.700 --> 00:07:13.860
T-inverse of b.

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And this should-- I don't know
which one your brain processes

00:07:16.990 --> 00:07:19.500
easier, but either of these,
when you when you take the

00:07:19.500 --> 00:07:21.120
composition of T with T-inverse,
you're going to be

00:07:21.120 --> 00:07:22.530
left with an a plus b.

00:07:22.530 --> 00:07:24.210
You take the composition of
T with T-inverse, you're

00:07:24.210 --> 00:07:25.180
left with an a.

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You take the composition of T
with T-inverse, you're just

00:07:27.500 --> 00:07:28.750
left with a b there.

00:07:28.750 --> 00:07:32.550
So in either case you get a
plus b-- when you evaluate

00:07:32.550 --> 00:07:36.350
either side of this expression
you'll get the vector a plus

00:07:36.350 --> 00:07:37.980
the vector b.

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Now what can we do?

00:07:41.370 --> 00:07:46.500
We know that T itself is a
linear transformation.

00:07:46.500 --> 00:07:49.190
And since T is a linear
transformation, we know that T

00:07:49.190 --> 00:07:58.490
applied to the sum of two
vectors is equal to T applied

00:07:58.490 --> 00:08:00.030
to each of those vectors
and summed up.

00:08:00.030 --> 00:08:01.360
Or we can take it
the other way.

00:08:01.360 --> 00:08:04.970
T applied to two separate
vectors-- so we call this one

00:08:04.970 --> 00:08:07.360
vector right here, and this
vector right here.

00:08:07.360 --> 00:08:10.540
So in this case I have a T
applied to one vector, and I'm

00:08:10.540 --> 00:08:12.700
summing it to a T applied
to another vector.

00:08:12.700 --> 00:08:17.790
So it's this right here, which
we know is equal to T applied

00:08:17.790 --> 00:08:20.480
to the sum of those
two vectors.

00:08:20.480 --> 00:08:24.294
So this is T applied to the
vector T-inverse of a-- let me

00:08:24.294 --> 00:08:33.610
write it here-- plus
T-inverse of b.

00:08:33.610 --> 00:08:36.409
It might look a little
convoluted, but all I'm saying

00:08:36.409 --> 00:08:39.630
is, this looks just like this.

00:08:39.630 --> 00:08:45.435
If you say that x is equal to
T-inverse of a, and if you say

00:08:45.435 --> 00:08:47.970
that y is equal to
T-inverse of b.

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So this looks just like that.

00:08:49.660 --> 00:08:52.730
It's going to be equal to the
transformation T applied to

00:08:52.730 --> 00:08:54.340
the sum of those two vectors.

00:08:54.340 --> 00:08:57.360
So it's going to equal the
transformation T applied to

00:08:57.360 --> 00:09:01.160
the inverse of a plus
T-inverse of b.

00:09:01.160 --> 00:09:04.130
I just use the fact that T
is linear to get here.

00:09:04.130 --> 00:09:04.960
Now what can I do?

00:09:04.960 --> 00:09:08.130
Let me let me simplify
everything that I've written

00:09:08.130 --> 00:09:08.840
right here.

00:09:08.840 --> 00:09:11.660
So I now have-- let
me rewrite this.

00:09:11.660 --> 00:09:14.160
This thing up here, which is
the same thing as this.

00:09:14.160 --> 00:09:23.290
T, the composition of T, with
T-inverse, applied to a plus b

00:09:23.290 --> 00:09:27.350
is equal to the composition--
or actually not the

00:09:27.350 --> 00:09:34.450
composition, just T-- applied to
two vectors, T-inverse of a

00:09:34.450 --> 00:09:38.680
plus T-inverse of vector b.

00:09:38.680 --> 00:09:40.700
That's what we've
gotten so far.

00:09:40.700 --> 00:09:44.220
Now we're very close to proving
that this condition is

00:09:44.220 --> 00:09:46.790
true for T-inverse, if we can
just get rid of these T's.

00:09:46.790 --> 00:09:50.740
Well the best way to get rid
of those T's is to take the

00:09:50.740 --> 00:09:53.080
composition with T-inverse
on both sides.

00:09:53.080 --> 00:09:56.180
Or take the T-inverse
transformation of both sides

00:09:56.180 --> 00:09:57.095
of this equation.

00:09:57.095 --> 00:09:58.910
So let's do that.

00:09:58.910 --> 00:10:02.300
So let's take T-inverse of this
side, T-inverse of that

00:10:02.300 --> 00:10:06.030
side, should be equal to
T-inverse of this side.

00:10:06.030 --> 00:10:07.620
Because these two things
are the same thing.

00:10:07.620 --> 00:10:09.950
So if you put the same thing
into a function, you should

00:10:09.950 --> 00:10:12.410
get the same value
on both sides.

00:10:12.410 --> 00:10:15.470
So what is this thing on
the left-hand side?

00:10:15.470 --> 00:10:15.970
What is this?

00:10:15.970 --> 00:10:18.640
This is the composition-- let me
write it this way-- this is

00:10:18.640 --> 00:10:24.650
the composition of T-inverse
with T, that part, applied to

00:10:24.650 --> 00:10:27.210
this thing right here.

00:10:27.210 --> 00:10:30.160
I'm just changing the
associativity of this--

00:10:30.160 --> 00:10:36.070
applied to T-inverse of the
vector a plus the vector b.

00:10:36.070 --> 00:10:38.640
That's what this left
hand side is.

00:10:38.640 --> 00:10:42.795
This part right here, T-inverse
of T of this, these

00:10:42.795 --> 00:10:45.090
first two steps I'm just writing
as a composition of

00:10:45.090 --> 00:10:50.240
T-inverse with T applied
to this right here.

00:10:50.240 --> 00:10:52.590
That right there is the same
thing as that right there.

00:10:52.590 --> 00:10:54.580
So that was another
way to write that.

00:10:54.580 --> 00:10:59.280
And so that is going to be equal
to the composition of

00:10:59.280 --> 00:11:02.730
T-inverse with T-- I'll write
that in the same color-- a

00:11:02.730 --> 00:11:05.330
composition of T-inverse
with T.

00:11:05.330 --> 00:11:08.470
That's this part right here,
which is very similar to that

00:11:08.470 --> 00:11:16.900
part right there-- of this stuff
right here, of T-inverse

00:11:16.900 --> 00:11:22.150
of a plus T-inverse
of the vector b.

00:11:22.150 --> 00:11:27.460
Now by definition of what
T-inverse is, what is this?

00:11:27.460 --> 00:11:30.480
This is the identity
transformation on our domain.

00:11:30.480 --> 00:11:33.910
This is the identity
transformation on Rn.

00:11:33.910 --> 00:11:37.690
This is also the identity
transformation on Rn.

00:11:37.690 --> 00:11:41.100
So if you apply the identity
transformation to anything,

00:11:41.100 --> 00:11:42.070
you're just going
to get anything.

00:11:42.070 --> 00:11:45.480
So this is going to be equal
to-- I'll do it on both sides

00:11:45.480 --> 00:11:49.380
of the equation-- this whole
expression on the left-hand

00:11:49.380 --> 00:11:54.360
side is going to simplify to the
T-inverse of the vectors a

00:11:54.360 --> 00:11:56.250
plus the vector b.

00:11:56.250 --> 00:11:57.530
And the right-hand side
is just going to

00:11:57.530 --> 00:11:59.190
simplify to this thing.

00:11:59.190 --> 00:12:01.050
Is equal to-- because this
is just the identity

00:12:01.050 --> 00:12:04.830
transformation-- so it's just
equal this one, T-inverse of

00:12:04.830 --> 00:12:09.660
the vector a plus T-inverse
of the vector b.

00:12:09.660 --> 00:12:13.460
And just like that, T-inverse
has met its first condition

00:12:13.460 --> 00:12:14.860
for being a linear
transformation.

00:12:17.840 --> 00:12:19.636
Now let's see if we can do
the second condition.

00:12:24.550 --> 00:12:26.010
Let's do the same
type of thing.

00:12:26.010 --> 00:12:32.850
Let's take the composition of
T with T-inverse, let's take

00:12:32.850 --> 00:12:38.740
the composition of that on some
vector, let's call it ca.

00:12:38.740 --> 00:12:39.850
Just like that.

00:12:39.850 --> 00:12:41.790
Well we know what this is equal
to, this is equal to the

00:12:41.790 --> 00:12:44.090
identity transformation on Rn.

00:12:44.090 --> 00:12:48.510
So this is just going
to be equal to ca.

00:12:48.510 --> 00:12:50.575
Now what is a equal to?

00:12:53.780 --> 00:12:57.070
What is this thing right there--
I'll write it on the

00:12:57.070 --> 00:12:59.710
side right here, let me do it
in an appropriate color.

00:12:59.710 --> 00:13:04.620
Or we could say that a, the
vector a is equal to the

00:13:04.620 --> 00:13:08.450
transformation T with the
composition of T with

00:13:08.450 --> 00:13:12.790
T-inverse applied
to the vector a.

00:13:12.790 --> 00:13:14.950
Because this is just the
identity transformation.

00:13:14.950 --> 00:13:17.840
So we can rewrite this
expression here as being equal

00:13:17.840 --> 00:13:27.410
to c times the composition of
T with T-inverse applied

00:13:27.410 --> 00:13:31.440
to my vector a.

00:13:31.440 --> 00:13:34.830
And maybe it might be nice to
rewrite it in this form

00:13:34.830 --> 00:13:36.620
instead of this composition
form.

00:13:36.620 --> 00:13:40.690
So this left expression we can
just write as T of the

00:13:40.690 --> 00:13:45.020
T-inverse of c times the
vector a-- all I did is

00:13:45.020 --> 00:13:49.520
rewrite this left-hand side this
way-- is equal to this

00:13:49.520 --> 00:13:50.510
green thing right here.

00:13:50.510 --> 00:13:51.620
Well I'll rewrite similarly.

00:13:51.620 --> 00:13:57.080
This is equal to c times the
transformation T applied to

00:13:57.080 --> 00:14:01.690
the transformation T-inverse
applied to a.

00:14:01.690 --> 00:14:06.400
This is by definition what
composition means.

00:14:06.400 --> 00:14:10.950
Now T is a linear
transformation.

00:14:10.950 --> 00:14:17.190
Which means that if you take c
time T times some vector, that

00:14:17.190 --> 00:14:22.050
is equivalent to T times c times
T applied to c times

00:14:22.050 --> 00:14:22.520
that vector.

00:14:22.520 --> 00:14:25.130
This is one of the conditions
in your transformation.

00:14:25.130 --> 00:14:27.980
So this is always going
to be the case with T.

00:14:27.980 --> 00:14:30.860
So if this is some vector
that T is applying to,

00:14:30.860 --> 00:14:31.890
this is some scalar.

00:14:31.890 --> 00:14:34.180
So this thing, because we
know that T is a linear

00:14:34.180 --> 00:14:39.680
transformation, we can rewrite
as being equal to T applied to

00:14:39.680 --> 00:14:44.110
the scalar c times T-inverse
applied to a.

00:14:47.200 --> 00:14:48.790
And now what can we do?

00:14:48.790 --> 00:14:51.560
Well let's apply the T-inverse
transformation to

00:14:51.560 --> 00:14:52.400
both sides of this.

00:14:52.400 --> 00:14:53.190
Let me rewrite it.

00:14:53.190 --> 00:15:02.580
On this side we get T of
T-inverse of ca is equal to T

00:15:02.580 --> 00:15:07.080
of c times T-inverse times a.

00:15:07.080 --> 00:15:08.500
That's what we have so far.

00:15:08.500 --> 00:15:09.850
But wouldn't it be nice
if we could get rid

00:15:09.850 --> 00:15:10.960
of these outer T's?

00:15:10.960 --> 00:15:13.980
And the best way to do that
is to take the T-inverse

00:15:13.980 --> 00:15:16.230
transformation of both sides.

00:15:16.230 --> 00:15:17.850
So let's do that.

00:15:17.850 --> 00:15:22.440
T-inverse-- let's take that of
both sides of this equation,

00:15:22.440 --> 00:15:25.850
T-inverse of both sides.

00:15:25.850 --> 00:15:28.300
And another way that this
could be written.

00:15:28.300 --> 00:15:31.130
This is equal into the
composition of T-inverse with

00:15:31.130 --> 00:15:38.090
T applied to T-inverse applied
to c times our vector a.

00:15:38.090 --> 00:15:41.050
This right here, I just decided
to keep writing it in

00:15:41.050 --> 00:15:43.140
this form, and I took these two
guys out and I wrote them

00:15:43.140 --> 00:15:44.560
as a composition.

00:15:44.560 --> 00:15:46.590
And this on the right-hand side,
you can do something

00:15:46.590 --> 00:15:47.610
very similar.

00:15:47.610 --> 00:15:49.630
You could say that this is equal
to the composition of

00:15:49.630 --> 00:15:54.820
T-inverse with T times--
or not times,

00:15:54.820 --> 00:15:56.380
let me be very careful.

00:15:56.380 --> 00:15:59.060
Taking this composition, this
transformation, and then

00:15:59.060 --> 00:16:06.680
taking that transformation
on c times the inverse

00:16:06.680 --> 00:16:08.390
transformation applied to a.

00:16:08.390 --> 00:16:10.660
Let me be very clear
what I did here.

00:16:10.660 --> 00:16:13.710
This thing right here is
this thing right here.

00:16:13.710 --> 00:16:15.710
This thing right here is
this thing right here.

00:16:15.710 --> 00:16:20.440
And I just rewrote this
composition this way.

00:16:20.440 --> 00:16:24.000
And the reason why I did this
is because we know this is

00:16:24.000 --> 00:16:28.060
just the identity transformation
on Rn, and this

00:16:28.060 --> 00:16:31.540
is just the identity
transformation on Rn.

00:16:31.540 --> 00:16:33.880
So the identity transformation
applied to anything is just

00:16:33.880 --> 00:16:34.940
that anything.

00:16:34.940 --> 00:16:40.360
So this equation simplifies to
the in T-inverse applied to c

00:16:40.360 --> 00:16:45.070
times some vector a, is equal
to this thing, c times

00:16:45.070 --> 00:16:47.660
T-inverse times some vector a.

00:16:47.660 --> 00:16:51.370
And just like that, we've met
our second condition for being

00:16:51.370 --> 00:16:54.630
a linear transformation.

00:16:54.630 --> 00:16:58.070
The first condition
was met up here.

00:16:58.070 --> 00:16:58.870
So now we know.

00:16:58.870 --> 00:17:02.130
And in both cases, we use the
fact that T was a linear

00:17:02.130 --> 00:17:05.630
transformation to get to the
result for T-inverse.

00:17:05.630 --> 00:17:13.880
So now we know that if T is a
linear transformation, and T

00:17:13.880 --> 00:17:25.710
is invertible, then T-inverse
is also a linear

00:17:25.710 --> 00:17:26.960
transformation.

00:17:30.220 --> 00:17:32.480
Which might seem like a little
nice thing to know, but that's

00:17:32.480 --> 00:17:33.730
actually a big thing to know.

00:17:33.730 --> 00:17:37.250
Because now we know that
T-inverse can be represented

00:17:37.250 --> 00:17:39.580
as a matrix vector product.

00:17:39.580 --> 00:17:44.270
That means that T-inverse
applied to some vector x could

00:17:44.270 --> 00:17:48.330
be represented as the product
of some matrix times x.

00:17:48.330 --> 00:17:50.090
And what we're going to do is,
we're going to call that

00:17:50.090 --> 00:17:56.230
matrix the matrix a-inverse.

00:17:56.230 --> 00:17:58.900
And I haven't defined as well
how do you construct this

00:17:58.900 --> 00:18:01.890
a-inverse matrix, but we
know that it exists.

00:18:01.890 --> 00:18:05.870
We know this exists now,
because T is a linear

00:18:05.870 --> 00:18:07.000
transformation.

00:18:07.000 --> 00:18:09.060
And we can take it even
a step further.

00:18:09.060 --> 00:18:14.090
We know by the definition of
invertibility that the

00:18:14.090 --> 00:18:19.710
composition of T-inverse with
T is equal to the identity

00:18:19.710 --> 00:18:21.320
transformation on Rn.

00:18:24.710 --> 00:18:26.280
Well what is a composition?

00:18:26.280 --> 00:18:30.870
We know that T, if we take--
let me put it this way.

00:18:30.870 --> 00:18:35.730
We know that T of x
is equal to Ax.

00:18:35.730 --> 00:18:40.710
So if we write T-inverse, the
composition of T-inverse with

00:18:40.710 --> 00:18:48.350
T applied to some vector x is
going to be equal to first, A

00:18:48.350 --> 00:18:51.380
being applied to x is going
to be equal to Ax, this

00:18:51.380 --> 00:18:53.960
a right here, Ax.

00:18:53.960 --> 00:18:57.300
And then you're going to apply
A inverse-x, you're going to

00:18:57.300 --> 00:19:00.140
apply this right here.

00:19:00.140 --> 00:19:03.190
And we got that this is the
equivalent to-- when you take

00:19:03.190 --> 00:19:06.810
the composition, it's equivalent
to, or your

00:19:06.810 --> 00:19:10.160
resulting transformation matrix
of two composition

00:19:10.160 --> 00:19:13.700
transformations is equal to this
matrix matrix product.

00:19:13.700 --> 00:19:14.840
We got that a long time ago.

00:19:14.840 --> 00:19:17.170
In fact that was the motivation
for how a matrix

00:19:17.170 --> 00:19:18.770
matrix product was defined.

00:19:18.770 --> 00:19:20.680
But what's interesting here is,
this composition is equal

00:19:20.680 --> 00:19:26.610
to that, but it's also equal to
the identity transformation

00:19:26.610 --> 00:19:30.170
on Rn applied to that vector
x, which is equal to the

00:19:30.170 --> 00:19:33.600
identity matrix applied to x.

00:19:33.600 --> 00:19:34.250
Right?

00:19:34.250 --> 00:19:37.380
That is the n by n matrix, so
when you multiply by anything,

00:19:37.380 --> 00:19:39.200
you get that anything again.

00:19:39.200 --> 00:19:41.740
So we get a very interesting
result.

00:19:41.740 --> 00:19:47.460
A-inverse times A has to be
equal to the identity matrix.

00:19:47.460 --> 00:19:51.350
A-inverse, or the matrix
transformation for T-inverse,

00:19:51.350 --> 00:19:54.410
when you multiply that with the
matrix transformation for

00:19:54.410 --> 00:19:59.580
T, you're going to get
the identity matrix.

00:19:59.580 --> 00:20:02.660
And the argument actually
holds both ways.

00:20:02.660 --> 00:20:05.640
So we know this is true, but
the other definition of an

00:20:05.640 --> 00:20:08.650
inverse, or invertibility, told
us that the composition

00:20:08.650 --> 00:20:12.390
of T with T-inverse is equal to
the identity transformation

00:20:12.390 --> 00:20:15.480
in our codomain, which
is also Rn.

00:20:15.480 --> 00:20:16.380
iRn.

00:20:16.380 --> 00:20:19.500
So by the exact same argument,
we know that when you go the

00:20:19.500 --> 00:20:22.430
other way, if you apply
T-inverse first and then you

00:20:22.430 --> 00:20:26.460
apply T-- so that's equivalent
of saying apply T-inverse

00:20:26.460 --> 00:20:31.150
first, and then you apply T
to some x vector, that's

00:20:31.150 --> 00:20:35.280
equivalent to multiplying that
x vector by the identity

00:20:35.280 --> 00:20:38.260
matrix, the n by n
identity matrix.

00:20:38.260 --> 00:20:40.500
Or you could say, you could
switch the order.

00:20:40.500 --> 00:20:45.590
A times A-inverse is also equal
to the identity matrix.

00:20:45.590 --> 00:20:47.710
Which is neat, because we
learned that matrix matrix

00:20:47.710 --> 00:20:50.140
products, when you switch the
order they don't normally

00:20:50.140 --> 00:20:51.260
always equal each other.

00:20:51.260 --> 00:20:54.140
But in the case of an invertible
matrix and its

00:20:54.140 --> 00:20:56.840
inverse, order doesn't matter.

00:20:56.840 --> 00:20:59.920
You can take A-inverse times A
and get the identity matrix,

00:20:59.920 --> 00:21:02.140
or you could take A times
A-inverse and get

00:21:02.140 --> 00:21:03.490
the identity matrix.

00:21:03.490 --> 00:21:05.910
Now we've gotten this far, the
next step is to actually

00:21:05.910 --> 00:21:07.220
figure out how do
you construct.

00:21:07.220 --> 00:21:09.010
We know that this thing exists,
we know that the

00:21:09.010 --> 00:21:11.350
inverse is a linear
transformation, that this

00:21:11.350 --> 00:21:12.220
matrix exists.

00:21:12.220 --> 00:21:14.360
We see this nice property,
that when you multiply it

00:21:14.360 --> 00:21:17.530
times the transformation
matrix you get

00:21:17.530 --> 00:21:18.940
the identity matrix.

00:21:18.940 --> 00:21:20.820
The next step is to actually
figure out how to

00:21:20.820 --> 00:21:23.330
figure this guy out.


Simplifying conditions for invertibility

WEBVTT
Kind: captions
Language: en

00:00:00.000 --> 00:00:03.560
Well the whole premise of the
last series of videos was

00:00:03.560 --> 00:00:06.610
really just trying to get at
trying to figure out whether

00:00:06.610 --> 00:00:10.080
some transformation T-- let's
have some transformation that

00:00:10.080 --> 00:00:17.040
is a mapping from, let's say
it's a mapping from Rn to Rm,

00:00:17.040 --> 00:00:19.100
--the whole question is,
is T invertible.

00:00:26.790 --> 00:00:29.110
And we showed several videos
ago that a function, and a

00:00:29.110 --> 00:00:31.900
transformation is really just a
function, that a function is

00:00:31.900 --> 00:00:34.400
invertible if it meets
two conditions.

00:00:34.400 --> 00:00:36.740
So invertible.

00:00:36.740 --> 00:00:38.840
So I don't have to keep writing
the word over again.

00:00:38.840 --> 00:00:40.010
You have to have
two conditions.

00:00:40.010 --> 00:00:43.980
It has to be onto or essentially
it has to map to

00:00:43.980 --> 00:00:48.210
every member of your codomain
and it also has to be

00:00:48.210 --> 00:00:50.540
one-to-one.

00:00:50.540 --> 00:00:53.060
Another way of saying one-to-one
is that every

00:00:53.060 --> 00:00:56.960
member of your codomain is
mapped to by at most one

00:00:56.960 --> 00:00:58.700
member of your domain.

00:00:58.700 --> 00:01:01.060
And we did several videos where
we thought well if we

00:01:01.060 --> 00:01:04.599
had a transformation, a linear
transformation, that's defined

00:01:04.599 --> 00:01:11.140
by a matrix, A, where this would
be an m-by-n matrix, we

00:01:11.140 --> 00:01:19.150
said that this is going to be
met if the rank-- This is only

00:01:19.150 --> 00:01:24.740
met if the rank of A is equal to
the number of rows in your

00:01:24.740 --> 00:01:27.710
transformation matrix
is equal to m.

00:01:27.710 --> 00:01:31.100
And in the last video I just
showed that this is only true

00:01:31.100 --> 00:01:34.360
if every one of your column
vectors are linearly

00:01:34.360 --> 00:01:37.200
independent or that they all
are basis vectors for your

00:01:37.200 --> 00:01:43.720
column space or that the rank
of your matrix it has to be

00:01:43.720 --> 00:01:45.320
equal to n.

00:01:45.320 --> 00:01:48.600
Now in order for something to
be invertible, in order for

00:01:48.600 --> 00:01:51.290
the transformation to be
invertible, both of these

00:01:51.290 --> 00:01:52.350
things have to be true.

00:01:52.350 --> 00:01:55.440
Your rank of A has to be equal
to m and your rank of A has to

00:01:55.440 --> 00:01:56.490
be equal to n.

00:01:56.490 --> 00:01:58.050
So in order to be invertible,
a couple of

00:01:58.050 --> 00:01:59.890
things have to happen.

00:01:59.890 --> 00:02:04.340
In order to be invertible your
rank of your transformation

00:02:04.340 --> 00:02:07.660
matrix has to be equal to m,
which has to be equal to n.

00:02:07.660 --> 00:02:10.340
So m has to be equal to n.

00:02:10.340 --> 00:02:11.610
So we have an interesting
condition.

00:02:11.610 --> 00:02:13.145
You have to have a
square matrix.

00:02:17.180 --> 00:02:20.790
Your matrix has to be n-by-n.

00:02:20.790 --> 00:02:22.030
That's what this implies.

00:02:22.030 --> 00:02:24.930
If both of these are true, then
m has to be equal n and

00:02:24.930 --> 00:02:26.480
you're dealing with
a square matrix.

00:02:26.480 --> 00:02:29.650
Even more, you're dealing with
the square matrix where every

00:02:29.650 --> 00:02:31.760
one of the columns are
linearly independent,

00:02:31.760 --> 00:02:34.660
so this is our A.

00:02:34.660 --> 00:02:35.750
A looks like this.

00:02:35.750 --> 00:02:41.200
A1, A2, all the way to An.

00:02:41.200 --> 00:02:44.140
Since the rank of A is equal to
n, and this is of course an

00:02:44.140 --> 00:02:45.280
n-by-n matrix.

00:02:45.280 --> 00:02:50.240
We just said that this has to
be the case because its rank

00:02:50.240 --> 00:02:52.470
has to be equal to m, which is
the number of rows, and its

00:02:52.470 --> 00:02:53.870
rank has to be equal to n,
which is the number of

00:02:53.870 --> 00:02:56.780
columns, so the rows and columns
have to be the same.

00:02:56.780 --> 00:02:59.010
But the fact that your rank
is equal to the number of

00:02:59.010 --> 00:03:03.250
columns, that means that all
of your column vectors are

00:03:03.250 --> 00:03:06.300
bases for your column space, or
that if you put them into

00:03:06.300 --> 00:03:14.500
reduced row echelon form, what
are you going to get?

00:03:14.500 --> 00:03:17.130
Well all of these guys are basis
vectors so they're all

00:03:17.130 --> 00:03:20.780
going to be associated with
pivot vectors or they're all

00:03:20.780 --> 00:03:22.500
going to be associated
with pivot columns.

00:03:22.500 --> 00:03:26.570
So this is going to be 1, 0,
bunch of 0's, and then you're

00:03:26.570 --> 00:03:29.180
going to have a 0, 1, a bunch
of 0's like this.

00:03:29.180 --> 00:03:40.690
They're all going to be
associated with pivot columns

00:03:40.690 --> 00:03:43.610
when you go into reduced
row echelon form.

00:03:43.610 --> 00:03:45.090
So all of them are
pivot columns.

00:03:45.090 --> 00:03:46.890
It's an n-by-n matrix.

00:03:46.890 --> 00:03:49.470
So what is an n-by-n
matrix where every

00:03:49.470 --> 00:03:51.570
column is a pivot column?

00:03:51.570 --> 00:03:53.120
What is an n-by-n matrix?

00:03:53.120 --> 00:03:54.870
Let me write this.

00:03:54.870 --> 00:03:55.970
So you have an n.

00:03:55.970 --> 00:04:02.710
So the reduced row echelon form
of A has to be equal to

00:04:02.710 --> 00:04:15.350
an n-by-n matrix, cause A is
n-by-n, where every column is

00:04:15.350 --> 00:04:17.510
a linearly independent
pivot column.

00:04:17.510 --> 00:04:20.870
And I mean by definition of
reduced row echelon form you

00:04:20.870 --> 00:04:23.570
can't have the same pivot
column twice where every

00:04:23.570 --> 00:04:28.460
column is a linearly independent
pivot column.

00:04:28.460 --> 00:04:30.450
It's a little bit redundant, but
I think you get the idea.

00:04:30.450 --> 00:04:33.550
So what is an n-by-n matrix
where every column is a

00:04:33.550 --> 00:04:35.560
linearly independent
pivot column?

00:04:35.560 --> 00:04:42.960
Well that is just a matrix that
has 1's down the diagonal

00:04:42.960 --> 00:04:44.215
and everything else is a 0.

00:04:46.890 --> 00:04:50.940
Or, you've seen this matrix
before, this is just an n-by-n

00:04:50.940 --> 00:04:56.410
identity matrix or the identity
matrix on n or on Rn.

00:04:56.410 --> 00:04:58.965
So if you multiply this matrix
times any member of Rn, you're

00:04:58.965 --> 00:05:01.350
just going to get that
matrix again.

00:05:01.350 --> 00:05:02.740
But this is interesting.

00:05:02.740 --> 00:05:06.440
We now have a pretty usable
condition for invertibility.

00:05:06.440 --> 00:05:12.360
We can say that the
transformation T that is a

00:05:12.360 --> 00:05:17.800
mapping from Rn to-- well it
has to map to the same

00:05:17.800 --> 00:05:21.300
dimension space so
from Rn to Rn.

00:05:21.300 --> 00:05:27.720
It's equal to some square matrix
n-by-n, it has to be an

00:05:27.720 --> 00:05:31.030
n-by-n matrix, times our
vectors in our domain.

00:05:31.030 --> 00:05:41.690
And it's only going to be
invertible if the reduced row

00:05:41.690 --> 00:05:47.550
echelon form of our
transformation matrix is equal

00:05:47.550 --> 00:05:50.620
to the identity matrix on n.

00:05:50.620 --> 00:05:54.590
I mean I could have written an
m here and I could've said

00:05:54.590 --> 00:05:58.160
this is an m-by-n matrix, but
the only way that this is

00:05:58.160 --> 00:06:01.700
going to be true is if
this is also an n and

00:06:01.700 --> 00:06:02.670
this is also an m.

00:06:02.670 --> 00:06:03.900
But maybe I could leave
them there.

00:06:03.900 --> 00:06:06.080
Let me leave those
m's there because

00:06:06.080 --> 00:06:07.700
that's the big takeaway.

00:06:07.700 --> 00:06:11.990
The big takeaway is that in
order for the transformation

00:06:11.990 --> 00:06:16.330
matrix to be invertible, the
only way it's invertible is if

00:06:16.330 --> 00:06:20.640
the reduced row echelon form of
our transformation matrix

00:06:20.640 --> 00:06:23.870
is equal to an n-by-n
identity matrix.

00:06:23.870 --> 00:06:27.410
The identity matrix is always
going to be n-by-n.

00:06:27.410 --> 00:06:28.510
So that's a big takeaway.

00:06:28.510 --> 00:06:31.880
Now we'll use that in the future
to actually solve for

00:06:31.880 --> 00:06:34.620
transformations or solve for
inverses of transformations.


Matrix condition for one-to-one trans

WEBVTT
Kind: captions
Language: en

00:00:00.530 --> 00:00:02.720
Let's say I have
some matrix A.

00:00:02.720 --> 00:00:06.090
If I'm trying to determine
the null space of A, I'm

00:00:06.090 --> 00:00:09.200
essentially asking -- I'm
essentially just asking, look

00:00:09.200 --> 00:00:14.740
if we set up the equation Ax is
equal to the 0 vector, the

00:00:14.740 --> 00:00:21.090
null space of A is all the x's
that satisfy this equation.

00:00:21.090 --> 00:00:29.800
All the x's that satisfy
that equation.

00:00:29.800 --> 00:00:32.370
Ax is equal to the
0 vector, or you

00:00:32.370 --> 00:00:33.680
could call it the system.

00:00:33.680 --> 00:00:35.930
And the way you would solve it
-- and we've done this many

00:00:35.930 --> 00:00:38.900
times, this was many videos
ago -- you would make an

00:00:38.900 --> 00:00:40.680
augmented matrix with this.

00:00:40.680 --> 00:00:42.510
So the augmented matrix
would look like that.

00:00:42.510 --> 00:00:45.320
And you would have the 0 vector
on the right hand side.

00:00:45.320 --> 00:00:48.530
And then you'd perform a bunch
of row operations to put the

00:00:48.530 --> 00:00:51.770
left hand side into reduced
row echelon form.

00:00:51.770 --> 00:00:54.070
So, you would do a bunch
of operations.

00:00:54.070 --> 00:00:57.040
The left hand side would go into
reduced row echelon form.

00:00:57.040 --> 00:01:00.950
Let's call that reduced
row echelon form of A.

00:01:00.950 --> 00:01:04.239
And then the right hand side
is just going to stay 0,

00:01:04.239 --> 00:01:06.100
because you performed the
same row operations.

00:01:06.100 --> 00:01:09.040
But when you perform those row
operations on 0, you just get

00:01:09.040 --> 00:01:10.850
the 0 vector right here.

00:01:10.850 --> 00:01:14.080
And then when you un-augment--
when you create the system

00:01:14.080 --> 00:01:16.840
back from this right here,
because these two systems are

00:01:16.840 --> 00:01:19.500
equivalent, you are essentially
going to have your

00:01:19.500 --> 00:01:21.320
solution set look something
like this.

00:01:21.320 --> 00:01:23.960
You're going to have your--
let me write it like this,

00:01:23.960 --> 00:01:25.840
your solution set is
going to be equal

00:01:25.840 --> 00:01:29.230
to some scalar multiple.

00:01:29.230 --> 00:01:31.870
Let's say that, of your free
variables, your free variables

00:01:31.870 --> 00:01:33.060
are going to be the
scalar multiples.

00:01:33.060 --> 00:01:34.820
And you've seen this
multiple times, so

00:01:34.820 --> 00:01:35.860
I'll stay fairly general.

00:01:35.860 --> 00:01:39.520
But it's going to be some
multiple times, let's say,

00:01:39.520 --> 00:01:45.880
vector 1 plus some other scalar
times vector 2, these

00:01:45.880 --> 00:01:49.200
scalars tend to be your free
variables-- times vector 2 all

00:01:49.200 --> 00:01:52.930
the way to, I don't
know, whatever, c

00:01:52.930 --> 00:01:55.600
times your nth vector.

00:01:55.600 --> 00:01:56.830
I'm just trying to
stay general.

00:01:56.830 --> 00:01:58.910
We haven't seen any examples
that had more than two or

00:01:58.910 --> 00:01:59.800
three vectors here.

00:01:59.800 --> 00:02:03.100
But this is what, essentially,
your null space is spanned by

00:02:03.100 --> 00:02:05.220
these vectors right there.

00:02:05.220 --> 00:02:08.110
You get an equation-- you get
a solution set that looks

00:02:08.110 --> 00:02:11.560
something like that, and you
call that your null space.

00:02:11.560 --> 00:02:13.220
We've done that multiple
times.

00:02:13.220 --> 00:02:15.340
Your null space is that,
so it's all the linear

00:02:15.340 --> 00:02:20.060
combinations, or it's the span,
of these little vectors

00:02:20.060 --> 00:02:20.690
that you get here.

00:02:20.690 --> 00:02:25.660
And 1 and 2, all the
way to n, n.

00:02:25.660 --> 00:02:26.670
This is nothing new.

00:02:26.670 --> 00:02:29.010
I'm just restating something
that we've seen multiple,

00:02:29.010 --> 00:02:29.790
multiple times.

00:02:29.790 --> 00:02:32.330
We actually did this in
the previous video.

00:02:32.330 --> 00:02:34.850
I just maybe never wrote
it exactly like this.

00:02:34.850 --> 00:02:37.110
But what about the case when
you're solving the

00:02:37.110 --> 00:02:38.760
inhomogeneous equation.

00:02:38.760 --> 00:02:42.090
So the inhomogeneous equation
looks like this.

00:02:42.090 --> 00:02:47.920
So if I want to solve Ax is
equal to b, I would do

00:02:47.920 --> 00:02:49.230
something very similar
to this.

00:02:49.230 --> 00:02:50.640
I will create an augmented
matrix.

00:02:50.640 --> 00:02:54.200
We'll have A on the left hand
side, and I'd put b on the

00:02:54.200 --> 00:02:57.010
right hand side, and I'll
perform a bunch of row

00:02:57.010 --> 00:03:01.006
operations to put A into reduced
row echelon form.

00:03:01.006 --> 00:03:02.750
So, let me do that.

00:03:02.750 --> 00:03:05.250
So this left hand side will
be the reduced row

00:03:05.250 --> 00:03:07.300
echelon form of A.

00:03:07.300 --> 00:03:09.650
And then the right hand side,
whatever operations I did on

00:03:09.650 --> 00:03:11.990
A, I have to do on
the entire row.

00:03:11.990 --> 00:03:14.040
So I'll also be doing
them to b.

00:03:14.040 --> 00:03:16.010
So I'll have some
new vector here.

00:03:16.010 --> 00:03:20.060
Maybe I'll call it the vector--
maybe I'll call that

00:03:20.060 --> 00:03:21.200
the vector b prime.

00:03:21.200 --> 00:03:22.640
It's going to be different
than b, but let's

00:03:22.640 --> 00:03:24.430
just call it b prime.

00:03:24.430 --> 00:03:28.100
And so when you when you go back
to your-- I guess when

00:03:28.100 --> 00:03:30.520
you go out of the augmented
matrix world and rewrite it as

00:03:30.520 --> 00:03:32.380
a system, and you solve for
it, and we did this in the

00:03:32.380 --> 00:03:35.870
last video-- you'll get
your solution set.

00:03:35.870 --> 00:03:39.345
Your solution set that satisfies
this is going to be,

00:03:39.345 --> 00:03:42.500
x is going to be equal to this
b prime, whatever this new

00:03:42.500 --> 00:03:47.890
vector is, this b prime plus
something that looks

00:03:47.890 --> 00:03:49.770
exactly like this.

00:03:49.770 --> 00:03:52.380
It looks exactly like that, in
fact I'll copy and paste it,

00:03:52.380 --> 00:03:56.130
it'll look exactly like this.

00:03:56.130 --> 00:04:01.640
Let me see if it did copy.

00:04:01.640 --> 00:04:03.600
Copy and paste it.

00:04:03.600 --> 00:04:07.680
Edit, copy, and let
me paste it.

00:04:07.680 --> 00:04:11.090
So it will look something that
looks exactly like that.

00:04:11.090 --> 00:04:17.700
And we said in the last video
that you can kind of-- given

00:04:17.700 --> 00:04:19.959
this you can kind of think of
the solution set to the

00:04:19.959 --> 00:04:24.430
inhomogeneous equation is
equivalent to some particular

00:04:24.430 --> 00:04:31.070
solution, let's call that x
particular, some particular

00:04:31.070 --> 00:04:35.490
solution, plus some member
of your null space.

00:04:35.490 --> 00:04:39.410
So you could say it's plus some
homogeneous solution.

00:04:39.410 --> 00:04:42.820
So if you just pick particular
values for a, b and c, all of

00:04:42.820 --> 00:04:45.520
the different multiples of the
vector that span your null

00:04:45.520 --> 00:04:46.940
space, you'll get some

00:04:46.940 --> 00:04:49.910
particular homogeneous solution.

00:04:49.910 --> 00:04:53.380
So what I implied in the last
video, and I didn't show it to

00:04:53.380 --> 00:04:57.725
you rigorously, is that any
solution to the inhomogeneous

00:04:57.725 --> 00:05:01.990
system-- let me write it this
way, any solution-- and do it

00:05:01.990 --> 00:05:09.740
in white-- that's not white--
any solution to the

00:05:09.740 --> 00:05:23.370
inhomogeneous system Ax is equal
to b, this is a claim I

00:05:23.370 --> 00:05:34.010
made, will take the form-- some
particular solution--

00:05:34.010 --> 00:05:35.830
that was this right here,
maybe I should do it in

00:05:35.830 --> 00:05:39.570
green-- this is this right
here-- When you do the reduced

00:05:39.570 --> 00:05:42.300
row echelon form, it becomes
that b prime vector.

00:05:42.300 --> 00:05:49.270
Plus some homogeneous solution,
so some member of

00:05:49.270 --> 00:05:51.580
the null space.

00:05:51.580 --> 00:05:53.860
Now I didn't prove it to
you, but I implied

00:05:53.860 --> 00:05:54.665
that this is the case.

00:05:54.665 --> 00:05:57.040
And what I want to in this video
is actually do a little

00:05:57.040 --> 00:05:59.790
bit of a more rigorous proof,
but it's actually fairly

00:05:59.790 --> 00:06:01.390
straightforward.

00:06:01.390 --> 00:06:06.040
So let's first of all verify
that this is a solution.

00:06:06.040 --> 00:06:08.050
So let's verify that
this is a solution.

00:06:08.050 --> 00:06:10.690
So let's just put this into
out original equation.

00:06:10.690 --> 00:06:14.000
Let's remember our original
equation was Ax is equal to b.

00:06:14.000 --> 00:06:15.250
So let's verify.

00:06:17.760 --> 00:06:19.950
Let me write it as a question.

00:06:19.950 --> 00:06:24.460
Is that particular solution,
plus some homogeneous

00:06:24.460 --> 00:06:32.290
solution, a solution to
Ax is equal to b?

00:06:32.290 --> 00:06:35.240
Well to do that you just put
that in the place of x.

00:06:35.240 --> 00:06:37.100
So let's try it out.

00:06:37.100 --> 00:06:41.230
So A times this guy right here,
times some particular

00:06:41.230 --> 00:06:46.540
solution, plus some homogeneous
solution is going

00:06:46.540 --> 00:06:50.050
to be equal to A times the
particular solution, plus A

00:06:50.050 --> 00:06:53.440
times some member of
my null space.

00:06:53.440 --> 00:06:54.980
And what is this equal to?

00:06:54.980 --> 00:06:58.280
That is going to
be equal to b.

00:06:58.280 --> 00:06:58.510
Right?

00:06:58.510 --> 00:07:00.290
We're saying that this
is a particular

00:07:00.290 --> 00:07:02.370
solution to this equation.

00:07:02.370 --> 00:07:05.240
That is going to be equal to b
and that this is going to be

00:07:05.240 --> 00:07:08.540
equal to the 0 vector because
this is a solution to our

00:07:08.540 --> 00:07:10.300
homogeneous equation.

00:07:10.300 --> 00:07:13.820
So this is going to be
equal to b plus 0, or

00:07:13.820 --> 00:07:15.050
it's equal to b.

00:07:15.050 --> 00:07:19.130
So A times this vector right
here is indeed equal to b.

00:07:19.130 --> 00:07:21.430
So this is a solution.

00:07:21.430 --> 00:07:23.750
This is a solution, yes.

00:07:23.750 --> 00:07:26.850
Now the next question is, does
every solution to the

00:07:26.850 --> 00:07:29.400
inhomogeneous system, or does
any solution to the

00:07:29.400 --> 00:07:32.860
inhomogeneous system,
take this form?

00:07:32.860 --> 00:07:48.810
So does any solution x, to Ax
equal to b, take the form x is

00:07:48.810 --> 00:07:53.200
equal to some particular
solution, plus a member of our

00:07:53.200 --> 00:07:56.940
null space, or plus a
homogeneous solution.

00:07:56.940 --> 00:08:00.700
So to do that let's take-- let's
test out what happens

00:08:00.700 --> 00:08:08.270
when we multiply the vector
A times x-- let me

00:08:08.270 --> 00:08:09.860
write it this way.

00:08:09.860 --> 00:08:20.460
let's say that x is any solution
to Ax is equal to b.

00:08:20.460 --> 00:08:21.790
Let's start off with that.

00:08:21.790 --> 00:08:26.800
And let's see what happens when
we take A times x minus

00:08:26.800 --> 00:08:31.130
some particular solution
to this.

00:08:31.130 --> 00:08:34.820
So when we distribute the matrix
vector product, you get

00:08:34.820 --> 00:08:39.940
A times our any solution
minus A times

00:08:39.940 --> 00:08:42.460
our particular solution.

00:08:42.460 --> 00:08:43.700
Now what is this going
to be equal to?

00:08:43.700 --> 00:08:46.180
We're saying that this is a
solution to Ax equals b.

00:08:46.180 --> 00:08:47.960
So this is going to
be equal to b.

00:08:47.960 --> 00:08:51.100
And of course, any particular
solution to this when you

00:08:51.100 --> 00:08:53.870
multiply it by A is also
going to be equal to b.

00:08:53.870 --> 00:08:56.310
So it's going to be b minus b,
so that's going to be equal to

00:08:56.310 --> 00:08:57.780
the 0 vector.

00:08:57.780 --> 00:09:03.700
Or another way to think about
it is, x, the vector x minus

00:09:03.700 --> 00:09:12.190
our particular solution is
a solution to A times

00:09:12.190 --> 00:09:14.180
x is equal to 0.

00:09:14.180 --> 00:09:18.350
Think about this if you take
this, in parentheses right

00:09:18.350 --> 00:09:20.890
here, and you put it right
there, and you multiply it

00:09:20.890 --> 00:09:22.880
times A you get the 0 vector.

00:09:22.880 --> 00:09:25.010
We just did that, you get the
0 vector, because when you

00:09:25.010 --> 00:09:27.320
multiply each of these guys
by A you get b and

00:09:27.320 --> 00:09:28.810
you get b minus b.

00:09:28.810 --> 00:09:30.740
And so that you get 0.

00:09:30.740 --> 00:09:35.990
So you can say that x minus--
so our any solution x minus

00:09:35.990 --> 00:09:51.520
the particular solution of x is
a member of our null space.

00:09:51.520 --> 00:09:51.760
Right?

00:09:51.760 --> 00:09:55.420
By definition our null space is
all of the x's that satisfy

00:09:55.420 --> 00:09:56.910
this equation.

00:09:56.910 --> 00:09:59.450
So, since its a member of our
null space, we can say that it

00:09:59.450 --> 00:10:03.010
is equal to-- so our any
solution minus our particular

00:10:03.010 --> 00:10:06.920
solution is equal to some member
of our null space.

00:10:06.920 --> 00:10:10.690
We could say that it is equal
to a homogeneous solution.

00:10:10.690 --> 00:10:11.870
There might be more than one.

00:10:11.870 --> 00:10:17.245
A homogeneous solution.

00:10:20.730 --> 00:10:24.830
Now, if we just add our
particular solution to both

00:10:24.830 --> 00:10:28.300
sides of this, we get that any
solution-- remember, we assume

00:10:28.300 --> 00:10:31.780
that x is any solution to this--
that any solution is

00:10:31.780 --> 00:10:34.420
equal to our homogeneous
solution, is equal to a

00:10:34.420 --> 00:10:40.120
homogeneous solution, plus a
particular solution-- or plus

00:10:40.120 --> 00:10:41.820
our particular solution.

00:10:41.820 --> 00:10:43.360
So we've proven it both ways.

00:10:43.360 --> 00:10:47.980
That this is a solution to our
inhomogeneous equation, and

00:10:47.980 --> 00:10:51.130
that any solution to our
inhomogeneous equation takes

00:10:51.130 --> 00:10:53.450
this form right here.

00:10:53.450 --> 00:10:57.940
Now why am I so concerned with
this, and I've been kind of

00:10:57.940 --> 00:11:00.910
fixated on this inhomogeneous
equation for some time.

00:11:00.910 --> 00:11:04.060
But we've been talking about the
notion of a transformation

00:11:04.060 --> 00:11:05.310
being one-to-one.

00:11:09.370 --> 00:11:11.850
That was one of the two
conditions for a

00:11:11.850 --> 00:11:14.200
transformation to
be invertible.

00:11:14.200 --> 00:11:19.180
Now to be one-to-one-- so let me
draw a transformation here.

00:11:19.180 --> 00:11:23.930
So let's say this is my domain,
X, and this is my

00:11:23.930 --> 00:11:28.910
co-domain right here, Y, and I
have a transformation that

00:11:28.910 --> 00:11:30.360
maps from X to Y.

00:11:33.630 --> 00:11:37.520
In order for T to be
one-to-one-- so I'll write it

00:11:37.520 --> 00:11:40.230
like this, one-to-one.

00:11:40.230 --> 00:11:42.830
In order for T to be one-to-one
that means for any

00:11:42.830 --> 00:11:50.600
b that you pick here, for any
b that is a member of our

00:11:50.600 --> 00:12:02.450
co-domain, there is at most
one solution to A times x.

00:12:02.450 --> 00:12:05.380
And I'm assuming that A is our
transformation matrix, so we

00:12:05.380 --> 00:12:09.300
can write our transformation T
as being equal to some matrix

00:12:09.300 --> 00:12:11.930
times our vector
in our domain.

00:12:11.930 --> 00:12:16.790
So this would be Ax if this is
x right here, so T would map

00:12:16.790 --> 00:12:18.830
from that to that right there.

00:12:18.830 --> 00:12:21.040
So in order for our
transformation to be

00:12:21.040 --> 00:12:24.450
one-to-one, that means if you
pick any b here, there has to

00:12:24.450 --> 00:12:28.300
be, at most, one solution
to Ax is equal to b.

00:12:28.300 --> 00:12:31.090
Or another way to say that is
that there is at most one guy

00:12:31.090 --> 00:12:33.750
that maps into that element
of our co-domain.

00:12:33.750 --> 00:12:34.790
There might be none.

00:12:34.790 --> 00:12:36.870
So there could be no solution
to this, but there has to be

00:12:36.870 --> 00:12:39.480
at most one solution.

00:12:39.480 --> 00:12:46.715
Now, we just said that any
solution to an inhomogeneous--

00:12:46.715 --> 00:12:53.610
let me write it in blue-- any
solution takes the form-- if

00:12:53.610 --> 00:12:54.310
there is a solution.

00:12:54.310 --> 00:12:56.350
So if there isn't a solution,
that's fine.

00:12:56.350 --> 00:12:58.270
That will still satisfy
one-to-one.

00:12:58.270 --> 00:13:01.050
But if there is a solution, any
solution is going to take

00:13:01.050 --> 00:13:09.500
the form x particular plus a
member of your null space.

00:13:09.500 --> 00:13:12.420
Where this guy right here is
a member of the null space.

00:13:12.420 --> 00:13:14.980
This thing right here just
applies to that guy.

00:13:14.980 --> 00:13:15.340
Right here.

00:13:15.340 --> 00:13:17.100
Any solution if they exist,
if there are no

00:13:17.100 --> 00:13:18.890
solutions, that's fine.

00:13:18.890 --> 00:13:20.740
You can still be one-to-one.

00:13:20.740 --> 00:13:23.850
But if you do have a solution
you can have at most one

00:13:23.850 --> 00:13:25.530
person that maps to it,
and any solution

00:13:25.530 --> 00:13:27.580
will take this form.

00:13:27.580 --> 00:13:28.840
I just showed you that.

00:13:28.840 --> 00:13:31.780
Now in order to be one-to-one,
this can only be one solution.

00:13:31.780 --> 00:13:33.610
The solution set can only
be one solution.

00:13:41.240 --> 00:13:45.010
We can only have one solution
here, right?

00:13:45.010 --> 00:13:46.120
What does that mean?

00:13:46.120 --> 00:13:49.580
That means that this guy
right here cannot be

00:13:49.580 --> 00:13:50.940
more than one vector.

00:13:50.940 --> 00:13:53.590
It just has to be one vector.

00:13:53.590 --> 00:13:57.000
There's only one particular
solution right there.

00:13:57.000 --> 00:14:01.500
But this guy right here has to
be-- for any solution set,

00:14:01.500 --> 00:14:03.200
depending on how you define
it, there's only one

00:14:03.200 --> 00:14:04.540
particular vector there.

00:14:04.540 --> 00:14:06.620
But this guy, the only way that
you're only going to have

00:14:06.620 --> 00:14:14.510
one solution is if your null
space is trivial, if it only

00:14:14.510 --> 00:14:16.240
contains the 0 vector.

00:14:16.240 --> 00:14:18.200
Your null space will
always, at minimum,

00:14:18.200 --> 00:14:19.120
contain the 0 vector.

00:14:19.120 --> 00:14:21.590
In the last video I think I,
just off the cuff, said that

00:14:21.590 --> 00:14:22.750
all your null space
has to be empty.

00:14:22.750 --> 00:14:26.270
But no, your null space will
always, by definition, by the

00:14:26.270 --> 00:14:29.480
fact that it is a subspace,
it will

00:14:29.480 --> 00:14:30.920
always contain a 0 vector.

00:14:30.920 --> 00:14:34.960
You can always multiply
A times 0 to get 0.

00:14:34.960 --> 00:14:36.930
So your null space will
always contain that.

00:14:36.930 --> 00:14:39.450
But in order to have only one
solution, your null space can

00:14:39.450 --> 00:14:43.340
only have the 0 vector so
that this can only be 0.

00:14:43.340 --> 00:14:46.040
And so that your only solution
is going to be the particular

00:14:46.040 --> 00:14:48.560
solution that you found,
depending on how you got

00:14:48.560 --> 00:14:51.220
there, but it's only going to
be your particular solution.

00:14:51.220 --> 00:14:54.760
So let me put this this way.

00:14:54.760 --> 00:15:04.110
So in order to be one-to-one
your null space of your

00:15:04.110 --> 00:15:06.830
transformation matrix
has to be trivial.

00:15:06.830 --> 00:15:11.280
It has to contain only
the 0 vector.

00:15:11.280 --> 00:15:13.590
Now we've covered this many,
many videos ago.

00:15:13.590 --> 00:15:16.910
What does it mean if your
null space only

00:15:16.910 --> 00:15:18.290
contains a trivial vector?

00:15:18.290 --> 00:15:19.270
Let me make this clear.

00:15:19.270 --> 00:15:23.340
So if your transformation vector
looks like this, A1,

00:15:23.340 --> 00:15:28.800
A2, all the way to An, and
you're multiplying it times

00:15:28.800 --> 00:15:33.650
x1, x2, all the way to xn, and
the null space is all of the

00:15:33.650 --> 00:15:36.385
x's that satisfy this equation,
0, and you are going

00:15:36.385 --> 00:15:39.210
to have m 0's right there.

00:15:39.210 --> 00:15:41.810
So if your null space is
trivial, and we're saying that

00:15:41.810 --> 00:15:45.170
that is a condition for you
to be one-to-one, for your

00:15:45.170 --> 00:15:47.510
transformation to be one-to-one,
the transformation

00:15:47.510 --> 00:15:50.510
that's specified
by this matrix.

00:15:50.510 --> 00:15:53.140
If your null space is trivial,
what does that mean?

00:15:53.140 --> 00:15:56.600
That means that the only
solution to-- another way of

00:15:56.600 --> 00:16:05.250
writing this is x1 times a1,
plus x2 times a2, all the way

00:16:05.250 --> 00:16:11.240
to xn times an, is equal
to the 0 vector.

00:16:11.240 --> 00:16:13.290
These are equivalent statements
right here.

00:16:13.290 --> 00:16:15.250
I just multiplied each of
these terms times these

00:16:15.250 --> 00:16:16.360
respective column vectors.

00:16:16.360 --> 00:16:17.760
These are the same thing.

00:16:17.760 --> 00:16:20.720
Now if you say that your null
space has to be equal to 0,

00:16:20.720 --> 00:16:23.270
you're saying that the only
solution to this equation

00:16:23.270 --> 00:16:28.270
right here, the only scalars
that satisfy this equation--

00:16:28.270 --> 00:16:32.950
oh sorry, these aren't-- let me
actually-- because I wrote

00:16:32.950 --> 00:16:37.010
the scalars as vectors-- so
this guy right here, this

00:16:37.010 --> 00:16:39.030
statement right here, is
equivalent to x1 times a1,

00:16:39.030 --> 00:16:45.680
plus x2 times a2, plus all the
way to xn times an, is equal

00:16:45.680 --> 00:16:47.710
to the 0 vector.

00:16:47.710 --> 00:16:50.075
Where the x1's through
xn's are scalars.

00:16:50.075 --> 00:16:52.680
Now, if we say the null space
is 0, we're saying the only

00:16:52.680 --> 00:16:56.890
way that this is satisfied is
if your x1 all the way to xn

00:16:56.890 --> 00:16:58.740
is equal to 0.

00:16:58.740 --> 00:17:02.700
And this means, this is our
definition actually, of linear

00:17:02.700 --> 00:17:03.350
independence.

00:17:03.350 --> 00:17:09.200
That means that a1-- so the null
space being 0 also means

00:17:09.200 --> 00:17:15.920
that your column vectors of a--
let me write it this way--

00:17:15.920 --> 00:17:23.310
it also means that a1, a2, all
the way through an, are

00:17:23.310 --> 00:17:26.060
linearly independent.

00:17:29.670 --> 00:17:30.930
Now what does that mean?

00:17:30.930 --> 00:17:35.950
If all of these guys are
linearly independent, what is

00:17:35.950 --> 00:17:38.230
going to be the basis for
your column space?

00:17:38.230 --> 00:17:40.660
Remember, the column
space is the span.

00:17:40.660 --> 00:17:47.510
The column space of a is equal
to the span of a1, a2,

00:17:47.510 --> 00:17:48.920
all the way to an.

00:17:48.920 --> 00:17:52.840
What we just said, if we're
dealing with the one-to-one,

00:17:52.840 --> 00:17:55.780
or one of the conditions, or the
condition to be one-to-one

00:17:55.780 --> 00:17:58.550
is that your null space
has to be 0, or only

00:17:58.550 --> 00:17:59.900
contain the 0 vector.

00:17:59.900 --> 00:18:01.970
If your null space contains the
0 vector, then all of your

00:18:01.970 --> 00:18:03.790
columns are linearly
independent.

00:18:03.790 --> 00:18:08.925
If all of these columns span
your column space, and they

00:18:08.925 --> 00:18:13.810
are linearly independent,
then they form a basis.

00:18:13.810 --> 00:18:27.430
So that means that a1, a2, all
the way to an, are a basis for

00:18:27.430 --> 00:18:29.520
our column space.

00:18:29.520 --> 00:18:32.880
And then that means if all of
our column vectors here are

00:18:32.880 --> 00:18:34.900
linearly independent, they
obviously span our column

00:18:34.900 --> 00:18:37.880
space by definition, and they
all our linearly independent,

00:18:37.880 --> 00:18:39.060
they form the basis.

00:18:39.060 --> 00:18:44.610
So the dimension of our basis,
so the dimension of our column

00:18:44.610 --> 00:18:48.910
space, that's essentially the
number of vectors you need to

00:18:48.910 --> 00:18:50.710
form the basis, is going
to be equal to n.

00:18:50.710 --> 00:18:52.032
We have n columns.

00:18:52.032 --> 00:18:53.930
So it's going to
be equal to n.

00:18:53.930 --> 00:18:58.790
Or another way to say it is that
the rank of your matrix

00:18:58.790 --> 00:19:00.990
is going to be equal to n.

00:19:00.990 --> 00:19:04.520
So now we have a condition for
something to be one-to-one.

00:19:04.520 --> 00:19:08.630
Something is going to be
one-to-one if and only if, the

00:19:08.630 --> 00:19:10.780
rank of your matrix
is equal to n.

00:19:10.780 --> 00:19:11.680
And you can go both ways.

00:19:11.680 --> 00:19:16.120
If you assume something is
one-to-one, then that means

00:19:16.120 --> 00:19:19.100
that it's null space here has to
only have the 0 vector, so

00:19:19.100 --> 00:19:20.690
it only has one solution.

00:19:20.690 --> 00:19:24.070
If it's null space only has the
0 vector, then that means

00:19:24.070 --> 00:19:26.000
it's columns are linearly
independent.

00:19:26.000 --> 00:19:28.790
Which means that they all
are part of the basis.

00:19:28.790 --> 00:19:30.960
Which means that you have
n basis vectors, or you

00:19:30.960 --> 00:19:32.060
have a rank of n.

00:19:32.060 --> 00:19:32.910
Let's go the other way.

00:19:32.910 --> 00:19:35.480
If you have a rank of n, that
means that all of these guys

00:19:35.480 --> 00:19:37.390
are linearly independent.

00:19:37.390 --> 00:19:39.470
If all of these guys are
linearly independent, then the

00:19:39.470 --> 00:19:43.240
null space is just
the 0 vector.

00:19:43.240 --> 00:19:45.920
The null space is just the 0
vector, this part of your

00:19:45.920 --> 00:19:47.460
solution disappears.

00:19:47.460 --> 00:19:49.410
And then you are only left
with one solution.

00:19:49.410 --> 00:19:50.360
So you're one-to-one.

00:19:50.360 --> 00:19:55.200
So you're one-to-one if and
only if the rank of your

00:19:55.200 --> 00:19:58.620
transformation matrix
is equal to n.


Exploring the solution set of Ax = b

WEBVTT
Kind: captions
Language: en

00:00:00.610 --> 00:00:02.770
Let's say I have some linear
transformation

00:00:02.770 --> 00:00:04.870
T from R2 to R2.

00:00:04.870 --> 00:00:09.700
So that is R2, and
then this is R2.

00:00:09.700 --> 00:00:14.770
T just maps from any member of
R2 to another member of R2,

00:00:14.770 --> 00:00:16.020
just like that.

00:00:16.020 --> 00:00:17.240
And I'm going to define T.

00:00:17.240 --> 00:00:18.400
It's a linear transformation.

00:00:18.400 --> 00:00:19.980
I'm going to define it.

00:00:19.980 --> 00:00:23.110
When I take the transformation
of some member of R2, it's

00:00:23.110 --> 00:00:25.820
equivalent to multiplying it by
this matrix, by the matrix

00:00:25.820 --> 00:00:29.560
1 minus 3, minus 1, 3.

00:00:29.560 --> 00:00:32.610
So just to kind of understand
this transformation a little

00:00:32.610 --> 00:00:34.790
bit more, let's think about all
of the values that it can

00:00:34.790 --> 00:00:36.360
take on in our codomain.

00:00:39.970 --> 00:00:46.720
So if we say that 1 minus 3,
minus 1, 3, times any vector

00:00:46.720 --> 00:00:51.390
in our domain-- so x1, x2-- it's
going to be equal to some

00:00:51.390 --> 00:00:53.580
other vector in our codomain.

00:00:53.580 --> 00:00:55.860
Let me call that vector b.

00:00:55.860 --> 00:00:57.795
B is a member of R2.

00:00:57.795 --> 00:01:02.360
So it's going to be
equal to b1, b2.

00:01:02.360 --> 00:01:05.250
All I'm doing here is I'm
essentially-- if I call this

00:01:05.250 --> 00:01:09.200
right here, I call that matrix
A-- I'm just trying to find

00:01:09.200 --> 00:01:13.080
out what are all the possible
values for Ax is equal to b?

00:01:13.080 --> 00:01:19.980
I'm trying to find out all of
the possible b's in this case.

00:01:23.040 --> 00:01:26.330
So if I were to solve this
equation for any particular b,

00:01:26.330 --> 00:01:27.810
what I would do is I
would put this in

00:01:27.810 --> 00:01:29.600
reduced row echelon form.

00:01:29.600 --> 00:01:32.290
Well, first I would make an
augmented matrix of it.

00:01:32.290 --> 00:01:36.700
So I would put a 1, a minus
3, a minus 1, and a 3.

00:01:36.700 --> 00:01:40.250
And I would augment it with
the member of our codomain

00:01:40.250 --> 00:01:42.320
we're trying to be equal to.

00:01:42.320 --> 00:01:45.720
So b1, b2.

00:01:45.720 --> 00:01:48.140
And then, I would put it in
reduced row echelon form.

00:01:48.140 --> 00:01:49.670
So how can we do that?

00:01:49.670 --> 00:01:51.800
How can we put it in reduced
row echelon form?

00:01:51.800 --> 00:01:53.420
I'll keep my first
row the same.

00:01:53.420 --> 00:01:56.630
1 minus 3, and then b1.

00:01:56.630 --> 00:01:59.350
And then my second row, I'll
replace it with my second row

00:01:59.350 --> 00:02:00.820
plus my first row.

00:02:00.820 --> 00:02:03.730
So minus 1 plus 1 is 0.

00:02:03.730 --> 00:02:06.440
3 plus minus 3 is 0.

00:02:06.440 --> 00:02:09.259
And then b2 plus b1 is--
well I can just write

00:02:09.259 --> 00:02:14.170
that as b1 plus b2.

00:02:14.170 --> 00:02:16.080
So this is only going
to have solutions.

00:02:16.080 --> 00:02:17.560
I have an interesting
situation here.

00:02:17.560 --> 00:02:18.540
You've seen it before.

00:02:18.540 --> 00:02:19.790
I have a row of 0's.

00:02:22.310 --> 00:02:25.140
And the only way that this
is going to actually have

00:02:25.140 --> 00:02:31.160
solutions is if this thing right
here is also going to be

00:02:31.160 --> 00:02:32.370
equal to 0.

00:02:32.370 --> 00:02:45.420
So the only members b that are
members of Rm that have

00:02:45.420 --> 00:02:51.730
solutions are the ones where
if you add up their two

00:02:51.730 --> 00:02:57.560
terms-- so let's say b is equal
to b1 and b2-- are the

00:02:57.560 --> 00:03:01.080
ones where the two terms added
up have to be to 0.

00:03:01.080 --> 00:03:04.850
b1 plus b2 have to
be equal to 0.

00:03:04.850 --> 00:03:07.990
Or another way you could write
it is that b2 has to be equal

00:03:07.990 --> 00:03:10.410
to minus b1.

00:03:10.410 --> 00:03:13.680
So if we were to actually draw
our codomain-- let's do it.

00:03:13.680 --> 00:03:16.650
We always stay in the abstract,
but sometimes it's

00:03:16.650 --> 00:03:18.640
useful to draw an
actual example.

00:03:18.640 --> 00:03:21.850
Let's say that our
codomain is R2.

00:03:21.850 --> 00:03:24.280
Let me draw some axes
right here.

00:03:24.280 --> 00:03:29.010
Let's say that this
is my b1-axis,

00:03:29.010 --> 00:03:30.830
and this is my b2-axis.

00:03:30.830 --> 00:03:32.380
I could have called that
x and y, but I

00:03:32.380 --> 00:03:33.900
called that b1 and b2.

00:03:33.900 --> 00:03:37.820
What are all of the members
of my codomain that have a

00:03:37.820 --> 00:03:40.000
solution, that have a mapping?

00:03:40.000 --> 00:03:44.390
Well, b2 has to be equal
to minus b1.

00:03:44.390 --> 00:03:45.900
So it's going to
look like this.

00:03:48.960 --> 00:03:51.860
It's just going to be a line
with a slope of negative 1.

00:03:51.860 --> 00:03:58.650
This is all the b's that
have a solution.

00:04:05.000 --> 00:04:06.950
Because if you're not
on this line.

00:04:06.950 --> 00:04:09.110
If you're a member of your
codomain-- this is the

00:04:09.110 --> 00:04:11.040
codomain right here, R2.

00:04:11.040 --> 00:04:13.920
R2 is also our domain, but let
me make it very clear that

00:04:13.920 --> 00:04:15.850
this is the codomain
that I've drawn.

00:04:15.850 --> 00:04:17.890
This is what we're
mapping into.

00:04:17.890 --> 00:04:21.170
It's very clear that if we're
not on this line, if you pick

00:04:21.170 --> 00:04:25.440
somebody whose two terms don't
add up to equalling 0, or

00:04:25.440 --> 00:04:27.840
aren't the negative of each
other, if you pick someone

00:04:27.840 --> 00:04:31.760
over here in our codomain, and
then you try to solve the

00:04:31.760 --> 00:04:36.780
equation over here, 0 is going
to equal some non-zero number

00:04:36.780 --> 00:04:39.750
here and you're not going
to have a solution.

00:04:39.750 --> 00:04:41.930
And we touched on this
in the last video.

00:04:41.930 --> 00:04:47.250
And so in this case we could
say that this right here is

00:04:47.250 --> 00:04:48.773
the image of our
transformation.

00:04:51.720 --> 00:04:53.320
Or even another way of
thinking about it.

00:04:59.520 --> 00:05:00.440
This is our codomain.

00:05:00.440 --> 00:05:01.690
Let me draw a domain.

00:05:06.430 --> 00:05:09.920
So if you take any member of
R2, and it's always mapping

00:05:09.920 --> 00:05:13.820
onto something onto that line,
clearly each point on that

00:05:13.820 --> 00:05:18.050
line is going to be mapped to
by more than one vector.

00:05:18.050 --> 00:05:20.840
So we're not dealing with
an onto transformation.

00:05:20.840 --> 00:05:22.420
We saw that in the last video.

00:05:22.420 --> 00:05:25.460
In order for something to be
onto, when you put it in

00:05:25.460 --> 00:05:29.750
reduced row echelon form,
you cannot have all 0's

00:05:29.750 --> 00:05:30.380
in one of the rows.

00:05:30.380 --> 00:05:32.800
Or another way to say it in
reduced row echelon form,

00:05:32.800 --> 00:05:36.820
every row has to have
a pivot entry.

00:05:36.820 --> 00:05:38.990
Well, let's focus
on the b's that

00:05:38.990 --> 00:05:40.460
actually do have a solution.

00:05:40.460 --> 00:05:43.190
So let's focus on these b's,
so that when you take a b1

00:05:43.190 --> 00:05:46.530
plus b2, it actually is going
to be equal to 0.

00:05:46.530 --> 00:05:49.790
So we could have the b, I don't
know, that could be the

00:05:49.790 --> 00:05:55.260
b5 minus 5.

00:05:55.260 --> 00:05:57.120
Obviously the b0, 0
is going to work.

00:05:57.120 --> 00:05:58.560
You could have 1, negative 1.

00:05:58.560 --> 00:05:59.820
Maybe that's right there.

00:05:59.820 --> 00:06:04.600
Let's focus on those for a
second and see how many

00:06:04.600 --> 00:06:07.030
members of our domain
map into them.

00:06:07.030 --> 00:06:10.390
So if we take this right here
and then we apply this

00:06:10.390 --> 00:06:13.960
equation up here, we only
have one constraint.

00:06:13.960 --> 00:06:15.650
We're assuming that this is
going to be equal to 0.

00:06:15.650 --> 00:06:18.340
So let's assume that
we're dealing with

00:06:18.340 --> 00:06:22.050
the b's in our image.

00:06:22.050 --> 00:06:24.890
Let's assume that we're dealing
with something where

00:06:24.890 --> 00:06:25.920
we can get a solution.

00:06:25.920 --> 00:06:28.890
That b1 plus b2 is equal to 0.

00:06:28.890 --> 00:06:30.210
Then what is our constraints?

00:06:30.210 --> 00:06:35.090
What will map to our vector
b that we're dealing with?

00:06:35.090 --> 00:06:38.710
If we just take this topic
equation right here, we have 1

00:06:38.710 --> 00:06:53.000
times x1-- let me switch
colors-- minus 3 times x2 is

00:06:53.000 --> 00:06:57.190
equal to b1.

00:06:57.190 --> 00:06:59.610
And then this row will give us
no constraints, because it's

00:06:59.610 --> 00:07:00.900
just going to be
a bunch of 0's.

00:07:00.900 --> 00:07:05.240
So this is the only constraint
for a member of our domain

00:07:05.240 --> 00:07:08.770
that will map to some particular
b now that we are

00:07:08.770 --> 00:07:11.750
picking, some particular
b that satisfies this

00:07:11.750 --> 00:07:12.560
constraint.

00:07:12.560 --> 00:07:14.820
So we could write this solution
set-- let me rewrite

00:07:14.820 --> 00:07:21.110
it as x1 is equal
to b1, plus 3x2.

00:07:21.110 --> 00:07:23.890
Or if we wanted to write the
entire solution set, it would

00:07:23.890 --> 00:07:26.280
look like this.

00:07:26.280 --> 00:07:39.960
x1, x2 is equal to the vector
b1, 0, plus-- x1 is b1, plus

00:07:39.960 --> 00:07:43.440
3, times x2, so plus
x2 times 3.

00:07:43.440 --> 00:07:45.910
And x2 is just going
to be equal to x2.

00:07:45.910 --> 00:07:47.260
That's a free variable.

00:07:47.260 --> 00:07:50.910
So x2 is equal to 0
plus x2, times 1.

00:07:50.910 --> 00:07:54.820
So what we're saying here is
this transformation just maps

00:07:54.820 --> 00:07:58.400
to this line here for all of the
vectors in our codomain,

00:07:58.400 --> 00:08:01.770
where their two entries
add up to each other.

00:08:01.770 --> 00:08:04.850
Let's assume that we actually
have one of those vectors.

00:08:04.850 --> 00:08:07.440
And so first of all, this is
definitely not an onto

00:08:07.440 --> 00:08:08.160
transformation.

00:08:08.160 --> 00:08:09.710
But let's assume that
we are dealing with

00:08:09.710 --> 00:08:10.580
one of those guys.

00:08:10.580 --> 00:08:13.660
So if we are picking a
particular one of those guys

00:08:13.660 --> 00:08:23.540
for a particular b-- let me
write this down-- that has a

00:08:23.540 --> 00:08:42.559
solution to Ax is equal to b,
the solution set will be equal

00:08:42.559 --> 00:08:43.690
to this thing right here.

00:08:43.690 --> 00:08:49.100
It'll be equal to x1, x2
is equal to the first

00:08:49.100 --> 00:08:50.200
entry of your b.

00:08:50.200 --> 00:08:57.020
b1, 0 plus x2, times
the vector 3,1.

00:08:57.020 --> 00:08:59.780
And if you think about this, if
you pick a particular b--

00:08:59.780 --> 00:09:03.610
so let's say we pick our-- let
me draw this out, because I

00:09:03.610 --> 00:09:07.450
think it's nice to
visualize it all.

00:09:07.450 --> 00:09:09.380
Actually, maybe I'll
draw it like this.

00:09:09.380 --> 00:09:11.250
I don't want to draw
our blobs anymore.

00:09:11.250 --> 00:09:13.250
So let me draw my axes.

00:09:13.250 --> 00:09:14.980
So my axes look like this.

00:09:14.980 --> 00:09:18.070
We know that the image of our
transformation is the line

00:09:18.070 --> 00:09:20.710
with a negative 1 slope, because
the two entries have

00:09:20.710 --> 00:09:22.510
to be equal to the negative
of each other.

00:09:22.510 --> 00:09:27.120
Let's pick a particular
b that has a solution.

00:09:27.120 --> 00:09:29.550
So let's say we pick
that b right there.

00:09:29.550 --> 00:09:31.710
In order for it to have a
solution, its entries have to

00:09:31.710 --> 00:09:32.830
be the negative of each other.

00:09:32.830 --> 00:09:38.480
So let's say that's the
entry 5, minus 5.

00:09:38.480 --> 00:09:40.660
That is our b.

00:09:40.660 --> 00:09:44.245
So what we just showed is that
the solution set, if we want

00:09:44.245 --> 00:09:47.060
to say, hey, what in our domain
maps to this guy?

00:09:47.060 --> 00:09:49.210
So let's think about what
in our domain maps

00:09:49.210 --> 00:09:50.460
to this right here.

00:09:52.880 --> 00:09:56.780
Let's think about what in our
domain maps to this point

00:09:56.780 --> 00:09:58.680
right here, to this
particular b.

00:09:58.680 --> 00:10:03.940
So that's going to be all of
the x's that satisfy Ax is

00:10:03.940 --> 00:10:07.425
equal to 5, minus 5.

00:10:13.050 --> 00:10:19.230
So x1, x2 is going to be equal
to b1, it's going to be equal

00:10:19.230 --> 00:10:26.910
to 5, 0 plus x2, plus any
scale or multiple of the

00:10:26.910 --> 00:10:28.420
vector 3, 1.

00:10:28.420 --> 00:10:31.780
So our solution set is going to
be, you take the vector 5,

00:10:31.780 --> 00:10:36.430
0-- so maybe the vector 5, 0,
specifies this position right

00:10:36.430 --> 00:10:40.420
here-- and then you're going to
add to it multiples of the

00:10:40.420 --> 00:10:41.320
vector 3, 1.

00:10:41.320 --> 00:10:45.770
The vector 3, 1 looks like this,
1, 2, 3 and you go up 1.

00:10:45.770 --> 00:10:47.980
The vector 3, 1 is going
to look like this.

00:10:51.110 --> 00:10:53.070
So if you add multiples of
this-- multiples of that could

00:10:53.070 --> 00:10:55.610
stretch out like that or could
go negative like that-- to

00:10:55.610 --> 00:10:59.440
this vector 5, 0, you're
essentially going to-- let me

00:10:59.440 --> 00:11:02.140
see if I can draw this neatly--
you're going to end

00:11:02.140 --> 00:11:08.090
up with a solution set that
looks like that right there.

00:11:08.090 --> 00:11:12.820
So if you pick a particular
b right there that has a

00:11:12.820 --> 00:11:17.240
solution, we just said that
everything on this line will

00:11:17.240 --> 00:11:19.620
map to that point in
our solution set.

00:11:23.810 --> 00:11:25.790
And in fact, if you picked
another point, let's say you

00:11:25.790 --> 00:11:29.050
picked the point minus 5, 5.

00:11:33.800 --> 00:11:36.700
Then the solution set that maps
to that would actually

00:11:36.700 --> 00:11:39.960
be-- this first term would
be the minus 5.

00:11:39.960 --> 00:11:40.940
It would be here.

00:11:40.940 --> 00:11:46.680
And all of these guys
would map to that.

00:11:46.680 --> 00:11:47.720
Well, this all interesting.

00:11:47.720 --> 00:11:49.680
I mean, we've been doing a
lot of abstract things.

00:11:49.680 --> 00:11:51.110
And I think it might be
satisfying that you're

00:11:51.110 --> 00:11:54.770
actually seeing something more
concrete in this example.

00:11:54.770 --> 00:11:56.370
But I'm doing all of
this for a reason.

00:11:56.370 --> 00:11:59.760
Because I want to understand
what the solution set is to a

00:11:59.760 --> 00:12:04.750
general non-homogeneous
equation like this.

00:12:04.750 --> 00:12:07.090
And to understand it a little
bit better, let's imagine what

00:12:07.090 --> 00:12:08.670
is the solution set
if I were to pick

00:12:08.670 --> 00:12:10.700
this guy, the 0 vector?

00:12:14.500 --> 00:12:15.910
Then what's going to be
the solution set?

00:12:19.660 --> 00:12:24.630
So if we say that Ax is equal to
0, then our solution set is

00:12:24.630 --> 00:12:33.050
going to the vector 0, 0,
plus x2, times 3, 1.

00:12:33.050 --> 00:12:34.350
So it's going to be what?

00:12:34.350 --> 00:12:37.540
This is just the 0 vector,
so it's going to be here.

00:12:37.540 --> 00:12:39.270
It's going to be here, and
it's going to just be

00:12:39.270 --> 00:12:41.410
multiples of 3, 1.

00:12:44.690 --> 00:12:47.020
It's going to look something
like that.

00:12:47.020 --> 00:12:47.770
What's this?

00:12:47.770 --> 00:12:51.470
What is the solution set to the
equation of Ax equal to 0.

00:12:51.470 --> 00:12:53.560
This is the null space.

00:12:53.560 --> 00:12:58.680
This, by definition, is
the null space of A.

00:13:02.150 --> 00:13:06.880
So notice-- and this is the big
takeaway from this video--

00:13:06.880 --> 00:13:10.630
that for any solution-- we're
picking b's that actually do

00:13:10.630 --> 00:13:12.650
have solutions, because we're
picking them on this line.

00:13:12.650 --> 00:13:16.320
We're picking them in the
image of our codomain.

00:13:16.320 --> 00:13:20.280
The solution set to any Ax is
equal to some b where b does

00:13:20.280 --> 00:13:23.800
have a solution, it's
essentially equal to a shifted

00:13:23.800 --> 00:13:28.450
version of the null set,
or the null space.

00:13:28.450 --> 00:13:31.290
This right here is
the null space.

00:13:31.290 --> 00:13:34.630
That right there is the null
space for any real number x2.

00:13:34.630 --> 00:13:36.900
Any scale or multiple of
3, 1 is the null space.

00:13:36.900 --> 00:13:37.960
I just showed it right there.

00:13:37.960 --> 00:13:39.300
It's going to be that.

00:13:39.300 --> 00:13:42.020
And so all of these other
solution sets are just some

00:13:42.020 --> 00:13:45.610
particular vector, some
x particular,

00:13:45.610 --> 00:13:49.050
plus the null space.

00:13:49.050 --> 00:13:52.300
Obviously, this vector by
itself would also be a

00:13:52.300 --> 00:13:55.860
solution to Ax is equal to b,
because you can just set x2 to

00:13:55.860 --> 00:13:57.120
be equal to 0.

00:13:57.120 --> 00:13:59.890
So in general-- and I haven't
proven this to you rigorously,

00:13:59.890 --> 00:14:02.950
but hopefully you kind of get
the intuition behind it.

00:14:02.950 --> 00:14:05.370
The solution-- and I'll do this
in the next video, just

00:14:05.370 --> 00:14:08.730
because I realize I'm running
long on time.

00:14:08.730 --> 00:14:21.300
Assuming Axb has a solution-- in
the example we just did, we

00:14:21.300 --> 00:14:24.070
can assume it has, if we pick
one of these points here.

00:14:24.070 --> 00:14:26.540
So assuming it has a solution,
if we pick a point off of our

00:14:26.540 --> 00:14:28.650
image, then we're not going
to have a solution.

00:14:28.650 --> 00:14:38.010
But assuming Axb has a solution,
the solution set is

00:14:38.010 --> 00:14:47.210
going to be equal to some
particular vector-- so you

00:14:47.210 --> 00:14:49.480
could just think of it as a set
with just one vector right

00:14:49.480 --> 00:14:53.110
there-- with, or combined with,
or the union of that

00:14:53.110 --> 00:15:02.620
set, with your null space of
this matrix right here.

00:15:02.620 --> 00:15:04.830
I haven't proven this to you
yet, but hopefully you get the

00:15:04.830 --> 00:15:06.410
intuition of why this is true.

00:15:06.410 --> 00:15:09.140
We just solved it for
particular cases

00:15:09.140 --> 00:15:10.100
that do have solutions.

00:15:10.100 --> 00:15:11.880
We say, hey, it's going
to take this form.

00:15:11.880 --> 00:15:13.740
And I just showed you
that this is the

00:15:13.740 --> 00:15:14.990
form of the null space.

00:15:18.100 --> 00:15:20.680
And the reason why we're doing
that is because we've been

00:15:20.680 --> 00:15:22.100
talking about invertibility.

00:15:22.100 --> 00:15:24.860
And in order to be invertible
you have to be

00:15:24.860 --> 00:15:27.520
onto, and 1 to 1.

00:15:27.520 --> 00:15:33.010
And for something to be 1 to
1, you have to have at most

00:15:33.010 --> 00:15:41.330
one solution that maps to
a particular vector.

00:15:41.330 --> 00:15:42.350
You might have none.

00:15:42.350 --> 00:15:44.350
But you might have,
at most, one.

00:15:44.350 --> 00:15:47.050
So in order to have, at most,
one solution-- and the

00:15:47.050 --> 00:15:49.950
solution set is always going to
be equal to this, so you're

00:15:49.950 --> 00:15:52.440
always going to have this
solution-- so in order to

00:15:52.440 --> 00:15:55.740
have, at most, one solution,
your null space can't have

00:15:55.740 --> 00:15:56.440
anything in it.

00:15:56.440 --> 00:15:58.315
Or it can just have
the zero vector.

00:16:01.290 --> 00:16:10.275
So that means that your null
space of A has to be trivial,

00:16:10.275 --> 00:16:20.430
or to be empty, or to just
have the zero vector.

00:16:20.430 --> 00:16:22.650
I'll do this a little bit more
rigorously in the next video,

00:16:22.650 --> 00:16:24.260
but I think when you do it
sometimes with the rigor, you

00:16:24.260 --> 00:16:26.540
don't necessarily get
the intuition.

00:16:26.540 --> 00:16:28.300
But this is going to be a very
interesting takeway, and I

00:16:28.300 --> 00:16:31.500
think you already understand
it's leading up to conditions

00:16:31.500 --> 00:16:32.750
for invertibility.


Simple Analysis of Cost per Job Saved from Stimulus

WEBVTT
Kind: captions
Language: en

00:00:00.810 --> 00:00:04.230
There's some news out today
about how many jobs, or at

00:00:04.230 --> 00:00:07.020
least a claim of how many jobs,
have been created so far

00:00:07.020 --> 00:00:08.045
from the stimulus plan.

00:00:08.045 --> 00:00:09.660
And at first, the numbers
look impressive.

00:00:09.660 --> 00:00:14.900
The administration said that
650,000 jobs were created from

00:00:14.900 --> 00:00:19.280
$150 billion of stimulus
spending so far.

00:00:19.280 --> 00:00:20.530
So let me write that down.

00:00:36.480 --> 00:00:39.870
So what does that turn out
to on a per-job basis?

00:00:39.870 --> 00:00:41.570
Let's see, let's get a
calculator out here.

00:00:52.660 --> 00:00:57.360
If I spent $150 billion to get
650,000 jobs-- let me just

00:00:57.360 --> 00:01:08.240
divide by 650,000-- I get
to $230,000 per job.

00:01:08.240 --> 00:01:17.100
So that is equal to $230,000
per job, which is

00:01:17.100 --> 00:01:18.500
claimed to be saved.

00:01:18.500 --> 00:01:21.520
So obviously, they're not just
outright paying the salary,

00:01:21.520 --> 00:01:23.880
these people were adding some
value to begin with.

00:01:23.880 --> 00:01:27.390
So these were clearly
very good jobs.

00:01:27.390 --> 00:01:31.820
So $230,000, I think it's fair
to say, is more than the

00:01:31.820 --> 00:01:33.310
majority of us make.

00:01:33.310 --> 00:01:35.140
So these are very good jobs.

00:01:38.660 --> 00:01:42.780
Now, there's a further claim
that at least another 350,000

00:01:42.780 --> 00:01:48.230
jobs were created from $189
billion of tax relief.

00:01:48.230 --> 00:01:51.470
We got $189 billion less to
the government and that

00:01:51.470 --> 00:01:53.705
created 350,000 jobs.

00:01:53.705 --> 00:01:55.350
So let's see what that
turns out to be.

00:02:09.320 --> 00:02:12.310
That's $189 billion
in tax relief.

00:02:15.360 --> 00:02:18.460
Let's see what that is
on a per job basis.

00:02:18.460 --> 00:02:20.410
Get the calculator out.

00:02:20.410 --> 00:02:36.870
And so, if we have $189 billion,
so you divide that by

00:02:36.870 --> 00:02:41.260
350,000 jobs.

00:02:41.260 --> 00:02:45.630
And you get $540,000 per job.

00:02:45.630 --> 00:02:55.350
So this is equal to $540,000 per
job, which are even better

00:02:55.350 --> 00:02:58.970
than the $230,000 jobs
that we were able to

00:02:58.970 --> 00:03:02.050
save with $150 billion.

00:03:02.050 --> 00:03:05.790
So clearly some very highly paid
people are being able to

00:03:05.790 --> 00:03:09.960
keep their jobs by virtue
of the stimulus package.