In our last video we stated some criteria
for a matrix to be diagonalizable
and it's boiled down to algebraic 
multiplicities being the same as
geometric multiplicities.
In this video we're going to examine
why that's really true.
So our big theorem was that we
had three conditions that were equivalent.
A matrix could be diagonalizable,
which is to say that there's a basis
for rn or cn that consists entirely of
eigenvectors of A
or that's equivalent to the sum of all the
geometric multiplicities adding up to n
where n is, we're talking about 
an n by n matrix.
And that's equivalent to every eigenvalue
having it's algebraic multiplicity
equal to a geometric multiplicity. 
And remember, the algebraic multiplicity
says what kind of root it is 
of the characteristic polynomial,
single root, double root, triple root.
Well the geometric multiplicity 
is the dimension of the eigenspace.
And we have a result leading up to 
that, and that said that the geometric
multiplicity, of course it's always 
at least one if you've got an eigenvalue
but it's never bigger than the 
algebraic multiplicity.
Why is it at least one?
Well if you have a root of the
characteristic polynomial then there's
a solution to a-λi*x=0
and that solution is an eigenvector so
you always have at least one eigenvector,
but the question is why can't the 
geometric multiplicity be bigger than
the algebraic?
Well, we're going to assume the
we have a certain geometric multiplicity
and we're going to compute the algebraic
and show that the algebraic has to be
at least as big as the geometric.
Now remember that when we figure out
the characteristic polynomial
we can work in whatever basis we want.
So let's suppose that we have an
eigenvalue λ_0, and let's call it's
geometric multiplicity k.
So that is to say there are k vectors,
b_1 through b_k, that form a basis
for the eigenspace. That means
they are all linearly independent.
So you can take those vectors and
then add a bunch of other vectors
to form a basis for all of rn or cn
as the case may be, depending on
whether we are using complex
eigenvalues or not.
And we are gonna compute
the matrix of A in that basis
and then we are gonna use that
matrix to compute the characteristics
polynomial. Now, the first k vectors
are eigenvectors with eigenvalue,
λ_0. The remaining vectors they might
be eigenvectors. They might not.
They are just vectors. So how do we
figure out the matrix? We have to
take Ab_1 and express it in the B 
basis, Ab_2 expressed in the B basis,
Ab_k expressed in the B basis.
And then A times all of the other,
the other vectors expressed in
the B basis. But Ab_1 is λ_1b_1.
So its coordinates in the B basis 
are just λ_0, 0, 0, 0, 0, 0, 0...
Ab_2 is λ_0b_2, so that's 0.
Its coordinates are 0, λ_0, et cetera.
Ab_3 is λ_0b_3. Ab_k is λ_0b_k.
So when we make this matrix,
in the upper left-hand corner,
we wind up with λ_0 times the identity.
And the lower left-hand corner
we wind up with 0.
Now what we get in the upper right
corner or in the lower right corner?
I have no idea. Because I don't know
anything about these vectors b_k+1,
b_k+2, b_k+3 and so on. They could
be anything. So let's give them a name.
We will just call whatever appears
In the upper right we will call that
sub-matrix C. Whatever appears
in the lower right, we call that
sub-matrix D. So with the upper-left,
we have a multiple identity.
And then C and D and down here
we have 0. Now, if you take λ times
the identity minus that, you get
(λ - λ_0)*I in the upper left hand corner.
0, -C and λ*I - D. And now you take
the determinant of that to get
the characteristic polynomial. 
Again, the determinant can be done
in any basis whatsoever and 
all bases give the same characteristic
polynomial. So you see that already
we pick up, k factors of λ - λ_0 from
over here and then we pick up
whatever the determinant gives you
down here. So you get a 
characteristic polynomial
that has k factors of λ - λ_0 and
might have some more factors
buried in here. You don't know. 
But it had has at least k factors.
So the algebraic multiplicity
has to be at least k which is to say,
at least as big as the geometric
multiplicity, which is what we want
to prove. So with that in hand,
we next observe the characteristic
polynomial is an nth order polynomial.
Every nth order polynomial has
exactly n roots counting multiplicity.
The sum of the algebraic multiplicity
is 0. And we know that the geometric
multiplicities are less than or equal
to the algebraic multiplicities. 
So the only way that the geometric
multiplicities can add up to n, 
sorry , the sum of algebraics add up
to n, not to 0. The only way that
the geometrics can add up to n,
is if the geometrics are all equal to
the algebraics. And some of the geometrics
can never be bigger than n.
So let's remember the three
criteria again. We said that something
Is diagonalizable if and only if the sum
of the geometric multiplicity is n,
which occurs if and only if
each geometric is equal to 
algebraic. And we have just shown
that these two conditions are the same.
Okay, to finish this off, we have to show
that these first two are the same.
Now if A is diagonalizable,
then you got a basis consisting of
eigenvectors. Well that means
that the sum of the dimension
of the eigenspaces is at least n.
On the other hand, the dimension,
the sum of the dimension
of the eigenspaces can't be bigger
than n. So it must be equal to n.
So if it's diagonalizable,
the sum of the geometric
multiplicities must be exactly n.
All that remains is to show that
if the sum of the geometric
multiplicity is n, then it's
diagonalizable. So to do that,
let's suppose that the sum of
the geometric multiplicity is n.
So you can write a basis for the first
eigenspace, a basis for the second
eigenspace and basis for the third
eigenspace. And you put all those
bases together, you get something
with n elements in it. And you'd like
this to be a basis for rn or cn.
Now if I give you a bunch of vectors,
how do you know whether they
form a basis, or n vectors in rn 
form a basis if they are linearly
independent. So we need to show
that these vectors are linearly
independent and claim they are.
So let's suppose it weren't,
we are gonna do a prove by
contradiction. Suppose they weren't
linearly independent. They are linearly
dependent. You could write 0 as
linear combination where the c_is
aren't all 0 and you could, you know,
might be able to do in several ways.
Let's do it in a way that involves
the fewest possible none-zero terms
that you can have.
So I rearrange the order of our vectors.
Let's assume that this combination
includes b_1.
Well, if we hit both sides of this 
equation with A - λ_1*I.
See A - λ_1*I acting on b_1 gives us
λ_1b_1. That's Ab_1 - λ_1b_1.
That's 0. It kills the first term. 
What it does to any other eigenvector
is it multiplies it by a constant.
So if you hit both sides of this equation
with A-λ_1*I, you get zero.
And then you kill the first term,
and the other terms survive.
All the terms of eigenvalue other than
λ_1 survived and you wind up with
a linear combination with fewer terms.
That's contradiction because we assume
that this was the combination of
the fewest possible terms.
And that's the end of the proof.
By contradiction, we got that
these were linearly independent.
Since they are linearly independent,
we have that the sum...
