In an earlier video, we talked about
diagonalizing Hermitian operators
and Hermitian matrices,
including real symmetric matrices.
And we discovered that they
had some very nice properties.
Well it turns out that unitary operators
have almost the same properties,
and they're almost as nice.
So let's suppose that we
have U is a unitary operator
on some finite-dimensional vector space.
Or if you prefer, you could think
of U as a unitary nxn matrix.
Then, the eigenvalues of U, they're
not real, they lie on the unit circle.
They can always be written as
e^iθ for some real value of θ.
They're associated with an angle.
Eigenvectors with different
eigenvalues are orthogonal.
Unitary matrices are
always diagonalizable;
we never have to worry
about power vectors.
And finally, you can always find an
orthonormal basis for your vector space
consisting of eigenvectors of U.
Now this is almost the same list of
properties we had for Hermitian matrices.
The only difference is that eigenvalues
of Hermitian matrices are real.
Eigenvalues of unitary matrices
lie on the unit circle.
So let's see why that is.
First let's suppose that
we have a unitary matrix.
And x is an eigenvector
with eigenvalue λ_1,
and y is an eigenvector
with eigenvalue λ_2.
So the inner product of x with itself
is the inner product of Ux with Ux
because U preserves inner products.
It's the inner product of λ_1x with λ_1x.
When you pull this λ_1 out of the
inner product, you get a λ_1-bar.
When you pull this one out, you get a λ_1.
And that tells us that λ_1-bar times λ_1,
in other words λ_1^2,
has to be 1.
So λ_1 has to be a unit complex number.
It doesn't have to be real,
it doesn't just have to be ±1,
it can be i or e^iπ/3.
It just has to be a complex
number of size 1.
It lives on the unit circle.
Now, how 'bout eigenvectors
with different eigenvalues?
Well the inner product of x with y
is the same thing as the
inner product of Ux with Uy,
because U preserves inner products.
Which is the inner product
of λ_1x with λ_2y,
which is λ_1-bar times λ_2
times the inner product of x with y...
But remember, λ_1 is
a unit complex number.
So λ_1-bar is the same thing as 1/λ_1.
So this is λ_2/λ_1 times the
inner product of x with y.
So the inner product of x with y
is a number other than 1
times the inner product of x with y.
And that means that the inner
product of x with y has to be 0.
You can put this on the
other side of the equation
and divide by 1 - λ_2/λ_1 if you wish.
But the upshot is that the
inner product has to be 0.
So here's an example.
Let's look at the matrix—this matrix.
Now you see, the first
column is a normal vector,
'cause (3/5)^2 + (4/5)^2 + 0^2 is 1.
The second column is a unit vector,
and the third column is a unit vector.
And they're orthogonal.
The inner product of these
two is -12/25 + 12/25, that's 0.
The inner product of the first and third
and the inner product of the second
and third are manifest to be 0.
So since the columns are orthonormal,
this must be a unitary matrix.
And in fact it is.
The eigenvalues are 3/5 ± 4/5i and i.
The eigenvectors are —you can
normalize it by dividing by √2—
, and .
Any two of these are orthogonal.
Where do the eigenvalues live?
They're on the unit circle.
Here is i, here is 3/5 + 4/5i,
here's 3/5 - 4/5i.
They're all on the unit circle.
Okay, so why are our unitary
matrices diagonalizable?
Now remember we had three
proofs of why Hermitian matrices
are diagonalizable,
and there are analogous proofs
for the unitary matrices.
The first is that there's no
such thing as power vectors,
or rather, any power vector
has to be an eigenvector.
So if the matrix weren't diagonalizable,
now you would have
a basis of power vectors
and you'd be able to find a
power vector of degree 2.
And you can scale it
so that its length is 1.
And then 1 would be the length of x^2,
so that would be the length of |(U^n)x|^2.
But if x is a power vector, then
we can expand (U^n)x
as (λ^n)x + nλ^(n-1)(U-λI)x.
And since it's not an eigenvector,
this term is not 0.
But that's an inner product
of this with itself.
As n gets bigger and bigger and
bigger, this term gets bigger
and bigger and bigger and bigger.
So the inner product with itself gets
bigger and bigger and bigger and bigger.
The inner product winds up being
n^2 times something nonzero
plus lower order terms.
In particular, it can't stay 1.
Any polynomial that has an n^2 in it
is not gonna be a constant.
That's a contradiction.
So you can never have a power vector
of degree 2 that's not an eigenvector,
and therefore it must be diagonalizable.
Same idea as for Hermitian matrices.
Orthogonality, this also works
like with Hermitian matrices.
You can approximate U by Uϵ
where Uϵ is unitary and has
distinct eigenvalues
and so is diagonalizable.
But then the eigenvectors
of Uϵ are orthogonal.
And then you take limits as ϵ goes to 0.
Each one of the eigenvectors of Uϵ has
a limit that becomes an eigenvector of U.
So you always have enough eigenvectors
of U to form an orthonormal basis.
Again, it's the same proof we
used for Hermitian matrices,
only with unitaries instead.
The last step is by induction.
We're gonna prove that unitary matrices
are diagonalizable by induction
on the size of the matrix.
It's obviously true for 1x1 matrices, so
let's suppose it's true for kxk matrices,
and we want to prove it
for (k+1)x(k+1) matrices.
So if we have a (k+1)x(k+1) matrix,
you can always find an eigenvector.
All matrices have at
least one eigenvector.
Let's call its eigenvalue λ.
And then you can pad this out
to form an orthonormal basis
that starts with the eigenvector and
has a bunch of other vectors.
Now, this first guy is an eigenvector;
these other guys can be anything.
They're not eigenvectors, they're just
any old vectors that, together with this,
make an orthonormal basis.
And then we look at the
matrix of U in the B basis.
Since B1 is an eigenvector, the first
column of U_B is .
And you notice, that's a unit vector.
|λ_1|^2 is 1,
+ 0^2 + 0^2 + 0^2 makes 1.
On the other hand, it's a unitary matrix,
so the first row has to be a
unit vector as well.
The only way the first row
can be a unit vector
is if the rest of the first row
is nothing but 0's.
So the first row has to be
(λ_1, 0, 0 , ..., 0),
and that means that our matrix
now is blocked diagonally.
And what's more is this portion
down here is unitary,
because the matrix of U† is the
transposed conjugate of this.
And so A† is A-inverse.
But we—By assumption,
you see this is a kxk matrix.
And this is 1x1.
We assume that every kxk matrix,
the unitary matrix, was diagonalizable.
So since A is diagonalizable,
U in the B basis is diagonalizable.
Which means that U is diagonalizable.
Again, it's morally the same proof
that we used for Hermitian operators.
Slight wrinkles about the rows,
but it's essentially the same proof.
