Applied linear operators and spectral methods/Lecture 3

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Review[edit | edit source]

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.

Adjoint operator[edit | edit source]

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.

Self-adjoint/Hermitian matrices[edit | edit source]

If we say that is self-adjoint, i.e., in any orthonomal basis, and the matrix is said to be Hermitian.

Anti-Hermitian matrices[edit | edit source]

A matrix is anti-Hermitian if

There is a close connection between Hermitian and anti-Hermitian matrices. Consider a matrix . Then

Jacobi Tridiagonalization[edit | edit source]

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 .

Spectral theory for matrices[edit | edit source]

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


Similarity transformation[edit | edit source]

The transformation

is called a similarity transformation. Two matrices are equivalent or similar is there is a similarity transformation between them.

Diagonalizing a matrix[edit | edit source]

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.

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