In the last video we talked about 
how you can use traces and determinants
to find eigenvalues and then eigenvectors
of matrices. We are gonna extend that
by using properties of scaling and 
identity. So let's start with a matrix
0 4 1 0 and you will notice that if you
multiply by 2 1, you get 4 2 which is
twice 2 1. In other words, 2 1 is an
eigenvector of 0 4 1 0 with eigenvalue 2.
So far so good.
But what if I instead of giving you
0 4 1 0, I give you 0 400 100 0.
Well you can multiply it out but
it's easier to notice that this is
just 100 times this matrix. And if you
take 100  A  x, that's gonna be
100Ax so that's gonna be 200x. You just
take this equation Ax = 2x and multiply
by 100.
So scaling a matrix by a factor of 100 
doesn't change the eigenvector. 2 1 used
to be an eigenvector. It's still an 
eigenvector.
All I did was scale the eigenvalue.
The matrix is 100 times bigger.
The eigenvalues are 100 times bigger.
Now let's look at this matrix.
This matrix is similar to this matrix
except it has 37 on the diagonal.
In other words, it's this matrix 
plus 37 times identity.
Now 37 times the identity times x
is just 37x and 100A * x was 200x.
So this is going to be 200x + 37x
and that's 237x.
By adding 37 times the identity
to the matrix, we didn't change
the eigenvector any, we just added
37 to the eigenvalue.
So let's summarize what we have
learned from this particular example.
If you take any matrix and scale it up
by a factor of c, you just scale up
the eigenvalues by a factor of c.
If you take a 100 times a matrix,
the eigenvalues are 100 times as big.
If you add a multiply of identity to
a matrix, you just add that number
to all the eigenvalues. If you add 
5 times the identify to the matrix,
you add 5 to all the eigenvalues.
And as for the eigenvectors, they don't
change. If you scale a matrix, or if you
add a multiple of the identify, the 
eigenvalues are exactly what they were
before. Now we can understand this
from the point of view of factorization.
If we can factorize a matrix as 
PDP^-1, with columns of P are
the eigenvectors and the diagonal
entries of D are the eigenvalues,
so D is λ_1, λ_n.
Then if you take c * A = P (cD) P^-1,
so the eigenvectors are the same.
P is the same. The eigenvalues
are factor c times larger.
If you add a multiple to the identity,
you can put the multiple of the identify
right in the middle because P times
the identity times P inverse
is just the identity.
So eigenvectors are the same.
Eigenvalues go up by whatever
constant you have here.
So let's work some examples.
So one of the simplest examples
of a matrix is 0 1 1 0.
The trace is 0. 0 + 0 is 0.
The determinant is -1.
0^2 - 1^2. That means the eigenvalues
have to be +/- 1. Add up to 0,
product is -1. The corresponding
eigenvectors are 1 1 and 1 -1
as you can check.
What about if you multiply that
by a constant b. 0 b b 0.
Well you should just think of that
as (0 1 1 0) * b. Instead of eigenvalues
being +/-1, they are now +/- b.
And the eigenvectors are the same
as they always were. Now let's add
a times the identity to that.
The eigenvalues of this matrix are
a bigger than the eigenvalues of
that matrix. In other words they are
a +/- b. And the eigenvectors,
they are still 1 1 and 1 -1. So whenever
you see a matrix of the form a b b a,
you should instantly know that 
the eigenvalues are a +/- b.
That the eigenvectors are 1 1 and 1 -1.
And that all comes from scaling.
You can also do this with complex
eigenvalues and eigenvectors.
We saw that the matrix 0 1 -1 0
had eigenvalues +/- i,
and eigenvectors +/-i 1. 
So if you scale that by a factor of b,
and add a times the identity,
you wind up with a +/- b*i,
and the eigenvectors are still +/-1 1.
Finally, I promised a 3 by 3 example
in the last video.
What are the eigenvalues of this matrix?
Big ugly 3 by 3 matrix.
Don't try to take this determinant.
It's gonna be a mess.
But you should think that it looks kind of
like the matrix 0 1 1 1 0 1 1 1 0.
We see that matrix before.
You probably have a homework problem
involving finding the eigenspaces.
The eigenvalues are 2 and -1.
-1 is a double root. So if you want,
we write as two -1, -1.
The trace is 2 + (-1) + (-1).
That's 0. The determinant is
2 times (-1) times (-1). That's 2.
Now once we know these eigenvalues,
what about that matrix?
Well it's just 5 times bigger.
Same matrix 5 times bigger.
So these eigenvalues have been 10, 
-5, and -5 and then we add 256 times
the identity to that. So we have to add
256 to all the eigenvalues.
So we get 266, 251 and 251.
That is, this is the eigenvector with
eigenvalues, with multiplicity 1,
251 is an eigenvalue with multiplicity 2.
And if you wanna check what the
eigenvectors are? Turns out
the eigenvectors here are 1 1 0,
and this is a two-dimensional eigenspace.
Its basis is -1 0 1 and -1 1 0.
And so here you have the eigen...
Sorry 1 1 1. Here the eigenvector is
1 1 1. And this eigenvector is -1 0 1,
and -1 1 0. It won't change.
