STEPHEN BEDDING:
In this recording,
I'm going to show how to
calculate eigenvectors for a 3
by 3 matrix.
I'm going to use the same matrix
whose eigenvalues are already
found in the previous recording.
The matrix was the
A, as written here.
Recall that the
eigenvalues we found
were 1, 2, and negative 1.
I've also written
down the combination
A minus lambda I
in terms of lambda.
You're going to
have to take lambda
equal to each of
eigenvalues in turn,
and look at three
different cases
to find the three
different eigenvectors.
Recall that the
eigenvectors-- I'll call them
v-- satisfy the
equation that, when
multiplied on the left by
the matrix A minus lambda I,
they must give 0.
I write my v as having
components v1, v2, v3,
and consider it to be a column.
The equations we
have to solve will
be equations for v1, v2, v3--
linear simultaneous equations.
We can solve those by
Gaussian elimination.
Let's write down
the basic matrix
for which we're going to
have to do the elimination.
It will look like this.
We'll have to do row
operations to try and find
values for v1, v2, and v3.
And we'll have to do them
three different times for each
of the different lambdas.
Let's start with
lambda equals 1.
We'll call it Case One.
I'll just have to put lambda
equals 1 in the matrix above.
That will give me 0, 1, negative
2, negative 1, 1, 1, and a 0,
1, negative 2.
Put the column of 0s on
the right, of course.
Elimination process is not
too hard for this matrix
because we've already
got some nice 0s.
I'd like a 1 to be in
the top left-hand corner.
So I'm going to multiply my
second row by negative 1,
and then swap it
with the first row.
Here's what we get.
Remember, I'm multiplying
the second row by negative 1,
but also changing the sign.
That will be 1,
negative 1, negative 1.
Got some 0s still.
And the rows have got swapped.
I suppose I probably ought
to say what I've done.
So let's see here.
Row two becomes negative row 2.
And then my notation
for swapping
is to use a squiggle like
this and put the rows in.
So R1 and R2 are swapped.
Now look.
We've got two rows the same.
The third row could be set 0.
R3 becomes R3 minus R2.
So that now gives us-- leaving
the first row the same.
So the second row.
The third row
completely disappears.
We're happy to see
that row of 0s there.
That means we can find a
non-trivial eigenvector.
We could now solve
the equations.
But I think I'm going to
do one more Gaussian step.
Look at the middle
column-- negative 1 and 1.
We could get rid of the negative
1 on top by adding row 2.
So let's do becomes-- row
1 becomes row 1 plus row 2.
That makes the result
1, 0, negative 3.
With 0s still down the right.
And everything else
remains the same.
Now I've got a nice form
for solving my equations.
What do the equations say?
The top row says v1
minus 3v3 equals 0.
And second row says
v2 minus 2v3 equals 0.
So we can express
everything in terms of v3.
v1 is 3v3, and v2 is 2v3.
Let's now write down
our eigenvector.
v must be v1 on top, v2 in
the middle, v3 at the bottom.
And we could pull the
v3 out as a factor.
And we've got 3, 2, 1.
And then, because it's only the
direction we care about and not
the length, we can choose v3.
And the sensible
choice here would
be to choose v3 just to be 1.
So my eigenvector fr
the eigenvalue lambda
equals 1 is just 3, 2, 1.
So this is the first
time we've done this.
So let's check it.
Let's go back and
write down the matrix
and multiply it by the v.
Here's the multiplication
we have to do.
We hope we're going to get
the answer 3, 2, 1 again,
because the eigenvalue is 1.
Let's do the matrix
multiplication.
So my answer is a column vector
formed from 3 plus 2 minus 2.
Negative 3 plus 4 plus 1.
And 2 minus 1.
That simplifies to 3, 2, 1.
Sure enough, we've
got the eigenvalue 1
times our vector, 3, 2, 1.
So that seems to have
worked very nicely.
Let's now repeat the process
for the second eigenvalue.
Second eigenvalue was
lambda equals negative 1.
We need to go back and
substitute that value of lambda
into our Gaussian
elimination equation.
That was back here.
This equation here.
Perhaps I should have
marked it as star.
No, it's not an equation.
It's a matrix.
OK.
I've got to put lambda
equals negative 1.
That's going to give me 2,
1, negative 2 across the top.
Let's write that down.
There's only going to be 0s
in the column to the right,
so I might as well put those in.
Now let's go back and
look at the middle row.
When lambda is negative 1 and
star, we have negative 1, 3, 1.
And then finally,
the bottom row.
Still in star, with lambda
equal to negative 1.
We get 0, 1, 0.
We have to do Gaussian
elimination on this one.
But actually, there's
not much to do.
Because that bottom row is
already rather friendly.
It's telling us v2 equals 0.
Now to put the other two rows.
With v2 equal to 0, we
can drop the v2 terms.
So the top row tells us
2v1 minus 2v3 equals 0.
And the middle row tells us
negative v1 plus v3 equals 0.
Those two equations
are actually the same.
The first one is just the second
one multiplied by negative 2.
So actually, we just
see that the answer
is v1 equals v3 with, at
the same time, v2 equals 0.
Let's put those together
into an eigenvector.
If v1 is v3 and v2 is
0, it'll look like this.
Once again, we could pull out
a factor of v3 and get 1, 0, 1.
Then we could let v3
be a number of choice.
I'm going to choose 1
again, for convenience.
So the eigenvector for lambda
equals negative 1 is just 1,
0, 1.
I'm going to leave you to
do the check this time.
You multiply the matrix
A by 1, 0, 1, and you
should get negative
1 times 1, 0, 1.
It does work.
But you try it.
I'm also going to leave
it as an exercise for you
to find the last eigenvector.
I'll write down the matrix that
you have to do the Gaussian
elimination on, and I'll
tell you the answer,
but I'd like you to try
and derive it yourself.
So here we are.
The eigenvalue is 2.
The Gaussian
elimination needs to be
done on the matrix
I've written here.
And then eigenvector
we find is 1, 3, 1.
But I'll leave that for
you to check, like I said.
