# Identifying Self-Adjoint Operators

1. Jan 14, 2010

### hbweb500

1. The problem statement, all variables and given/known data

If A has eigenvalues 0 and 1, corresponding to the eigenvectors (1,2) and (2, -1), how can one tell in advance that A is self-adjoint and real.

2. Relevant equations

e=m^2

3. The attempt at a solution

I can show that A is real: it has real orthogonal eigenvectors and real eigenvalues which form a basis of R2, so any real vector is transformed by A to another vector in R2. It follows that all of the components of A are real.

Showing that the matrix is self-adjoint, however, is trickier for me. I know that eigenvectors corresponding to distinct eigenvalues of a self-adjoint matrix are orthogonal, and clearly these eigenvectors are orthogonal. However, I don't think is is a sufficient condition. If a matrix has two orthogonal eigenvectors, surely that doesn't mean it is self-adjoint, right?

So how can you know at a glance (that is, without reconstructing A by finding the unitary operators formed by taking its eigenvectors as columns), that A is self-adjoint?

2. Jan 14, 2010

### rochfor1

What can you say about the diagonal matrix of eigenvalues?
What can you say about any conjugate of a self-adjoint matrix?

3. Jan 14, 2010

### Matterwave

Perhaps it'd be easier if you noted that a self-adjoint matrix which is real, is simply a symmetric matrix.

4. Jan 14, 2010

### hbweb500

I'm not quite getting why these are relevant. The conjugate of a self-adjoint matrix A is just A transposed. Not sure what is so interesting about the diagonal matrix of eigenvalues...

5. Jan 14, 2010

### rochfor1

Write down the diagonal matrix of eigenvalues in this situation. Write down its adjoint. Can you see a relationship?

6. Jan 15, 2010

### vela

Staff Emeritus
Think about how the original matrix A and the diagonal matrix of eigenvalues are related.

Edit: Oh, I missed you had written this:
Construct the matrix with the eigenvectors as columns. Do you notice anything about it?

7. Jan 15, 2010

### hbweb500

Bam!, I think I have it.

A is diagonalizable, so...

$$A = U^\ast \Lambda U$$

And so:

$$A^\ast = U^\ast \Lambda ^\ast U$$

But the eigenvalues are all real, so big lambda is self-adjoint, and we can say that:

$$A^\ast = U^\ast \Lambda U = A$$

I think that does it. Cool beans.