# 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.

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

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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.

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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).

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