# What does multiplying a matrix by its transpose have to do with spectral theorem? [closed]

What does multiplying a matrix by its transpose have to do with spectral theorem? I basically am trying to understand what this would mean with regards to spectra of waves.

I think it give you a diagonal matrix, but I'm not sure how it relates to spectral theory.

-

## closed as not a real question by Andreas Blass, Bill Johnson, Qiaochu Yuan, Michael Renardy, Pietro MajerNov 8 '12 at 19:47

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

The spectral theorem is for normal operators (in particular, hermitian, or real symmetric, matrices). If $A$ is a real matrix, $A A^T$ is a real symmetric matrix.

-

In the finite dimensional case the Spectral Theorem says that we can decompose a self-adjoint operator into a sum of projection operators: if $A$ is self-adjoint then we can write

$A=\lambda_{1}P_{1}+\cdots +\lambda_{r}P_{r}$

where the $\lambda$'s are the (necessarily real) eigenvalues of $A$ and the $P$'s are orthogonal projection onto the corresponding eigenspaces. If we choose an orthonormal basis of each eigenspace and let $U_{i}$ be the matrix whose columns are this eigenbasis, then we have $P_{i}=U_{i}U_{i}^{\ast}$.

As mentioned in the answer by Robert Israel, given any $A$ we always have $AA^{\ast}$ is self-adjoint so the discussion above holds. In particular, we can determine an orthonormal basis for which the operator determined by $A$ is a diagonal matrix real entries on the diagonal (ie, $A$ is (orthogonally) diagonalisable).

This material can be found in any good linear algebra textbook (my personal favourite is by Hoffmann & Kunze).

-