# Showing it is orthogonally diagonalizable

1. Mar 20, 2008

Suppose that the real matrices A and B are orthogonally diagonalizable and AB=BA. Show that AB is orthogonally diagonalizable.

I know that orthogonally diagonalizable means that you can find an orthogonal matrix Q and a Diagonal matrix D so Q^TAQ=D, A=QDQ^T.

I am aware of the Real Spectral Theorem which states that "A real (mxn)-matrix A is orthogonally diagonalizable if and only if A is symmetric"

I got a hint saying I am suppose to use the Real Spectral Theorem twice to show it. But I am still unsure as to how to do this.

2. Mar 20, 2008

### transgalactic

i am not femiliar with the termin "orthogonally diagonalizable "

in order to proove that its diagonazable you need to proove
that the eigenvectors are independant

orthogonal meens perpandicular
so i think you should take the columns of the matrix
and if the multiplication of each vector by another equals to zero
then its orthogonal

3. Mar 20, 2008

### StatusX

Use the fact that commuting matrices are simultaneously diagonalizable.

Specifically, since B commutes with A, if v is an eigenvector of A with eigenvalue a, then A(Bv)=B(Av)=B(av)=a(Bv), so Bv is another eigenvector of A with the same eigenvalue. Now, if the eigenspace corresponding to the eigenvalue a is one dimensional, this means Bv must be a multiple of v, ie, v is also an eigenvalue of B. If all the eigenspaces of A are one dimensional, then A and B have exactly the same eigenvectors, and so they are diagonalized by the same matrix Q (since the columns of this matrix are precisely the eigenvectors of the matrix being diagonalized).

I'll let you finish the argument and work out what happens when the eigenvalues are degenerate (ie, when some eigenspaces are more than one dimensional).

Last edited: Mar 20, 2008