### Eigenvalues and eigenvectors of a projector

**Exercise 1**. Find eigenvalues of a projector.

**Solution**. We know that a projector doesn't change elements from its image: for all This means that is an eigenvalue of Moreover, if is any orthonormal system in each of is an eigenvector of corresponding to the eigenvalue

Since maps to zero all elements from the null space is another eigenvalue. If is any orthonormal system in each of is an eigenvector of corresponding to the eigenvalue

A projector cannot have eigenvalues other than and This is proved as follows. Suppose with some nonzero Applying to both sides of this equation, we get It follows that and (because ) The last equation has only two roots: and

We have because is an orthogonal sum of and . Combining the systems we get an orthonormal basis in consisting of eigenvectors of .

### Trace of a projector

Recall that for a square matrix, its trace is defined as the sum of its diagonal elements.

**Exercise 2**. Prove that if both products and are square. It is convenient to call this property **trace-commuting** (we know that in general matrices do not commute).

**Proof**. Assume that is of size and is of size For both products we need only to find the diagonal elements:

All we have to do is change the order of summation:

**Exercise 3**. Find the trace of a projector.

**Solution**. In Exercise 1 we established that the projector has eigenvalues and eigenvalues is symmetric, so in its diagonal representation there are unities and zeros on the diagonal of the diagonal matrix By Exercise 2

.

## Leave a Reply

You must be logged in to post a comment.