In the last lecture we talked about norms in inner product spaces. The
induced norm was defined as
We also talked about orthonomal bases and biorthonormal bases. The biorthonormal
bases may be thought of as dual bases in the sense that covariant and
contravariant vector bases are dual.
The last thing we talked about was the idea of a linear operator. Recall that
where the summation is on the first index.
In this lecture we will learn about adjoint operators, Jacobi tridiagonalization,
and a bit about the spectral theory of matrices.
Assume that we have a vector space with an orthonormal basis. Then
One specific matrix connected with is the Hermitian conjugate matrix.
This matrix is defined as
The linear operator connected with the Hermitian matrix is called the
adjoint operator and is defined as
Therefore,
and
More generally, if
then
Since the above relation does not involve the basis we see that the adjoint operator is also basis independent.
If we say that is self-adjoint, i.e.,
in any orthonomal basis, and the matrix
is said to be Hermitian.
A matrix is anti-Hermitian if
There is a close connection between Hermitian and anti-Hermitian matrices.
Consider a matrix . Then
Let be self-adjoint and suppose that we want to solve
where is constant. We expect that
If is "sufficiently" small, then
This suggest that the solution should be in the subspace spanned by
.
Let us apply the Gram-Schmidt orthogonalization procedure where
Then we have
This is clearly a linear combination of .
Therefore, is a linear combination of
. This
is the same as saying that is a linear combination of
.
Therefore,
Now,
But the self-adjointeness of implies that
So is or . This is equivalent to expressing
the operator as a tridiagonal matrix which has the form
In general, the matrix can be represented in block tridiagonal form.
Another consequence of the Gram-Schmidt orthogonalization is that
Lemma:
Every finite dimensional inner-product space has an orthonormal basis.
Proof:
The proof is trivial. Just use Gram-Schmidt on any basis for that space and
normalize.
A corollary of this is the following theorem.
Theorem:
Every finite dimensional inner product space is complete.
Recall that a space is complete is the limit of any Cauchy sequence from a
subspace of that space must lie within that subspace.
Proof:
Let be a Cauchy sequence of elements in the subspace with
. Also let be an
orthonormal basis for the subspace .
Then
where
By the Schwarz inequality
Therefore,
But the ~s are just numbers. So, for fixed , is
a Cauchy sequence in (or ) and so converges to
a number as , i.e.,
which is is the subspace .
Suppose is expressed in coordinates relative to some basis
, i.e.,
Then
So implies that
Now let us try to see the effect of a change to basis to a new basis
with
For the new basis to be linearly independent, should be invertible so
that
Now,
Hence
Similarly,
Therefore
So we have
In matrix form,
where the objects here are not operators or vectors but rather the matrices and
vectors representing them. They are therefore basis dependent.
In other words, the matrix equation
The transformation
is called a similarity transformation. Two matrices are equivalent
or similar is there is a similarity transformation between them.
Suppose we want to find a similarity transformation which makes diagonal,
i.e.,
Then,
Let us write (which is a matrix) in terms of its columns
Then,
i.e.,
The pair is said to be an eigenvalue pair if where is an eigenvector and is an eigenvalue.
Since this means that is an
eigenvalue if and only if
The quantity on the left hand side is called the characteristic polynomial
and has roots (counting multiplicities).
In there is always one root. For that root
is singular, i.e., there always exists at least one eigenvector.
We will delve a bit more into the spectral theory of matrices in the next
lecture.