Suppose that A is a normal matrix with real eigenvalues Show

Suppose that A is a normal matrix with real eigenvalues. Show that A must be Hermitian.

By definition, a complex matrix A is normal if

A^* A = A A^* ,

where A^* is the conjugae transpose of A.

Let all the eigenvalues of A are real.

To prove:

A is Hermitian. i.e. A^* = A.

Proof:

Given:

A is normal.

By theorem:

A matrix A is normal if and only if it is orthogonally diagonalized.

That is:

We can find an orthogonal matrix U, such that:

1. U U^* = U^* U = I, where I is the identity matrix and

2. A = U^* D U, (1)

where D is a diagonal matrix, containing the eigenvalues of A in the diagonal.

Given:

The eigenvalues of A are real.

So,

D^* = D. (2)

From (1):

A^* = (U^* D U)^*

= U^* D^* U

= U^* D U, using (2)

= A.

Thus, we get:

A^* = A,

which proves that A must be Hermitian.