Vtyjkyuk

Let $ A $ be a real matrix. Prove that $ A cdot A^t $ is diagonaziable.

This is the only info given. Now I understand that $ A cdot A^t $ is a symmetric matrix because

$ [AA^t]_ij = [A]_i^r cdot [A^t]_j^c = [A]^r_i cdot [A]^r_j = [A^t]_i^c cdot [A]^r_j = [A]_j^r cdot [A^t]_i^c = [AA^t]_ji $

and I found this document where it says symmetric matrices has $ n$ eigenvalues (not necessarily distinct) but how can I know if there is an invertible matrix $ P $ that diagonalize $ A $ ?

Thanks

$A.A^t$ being symmetric, it is diagonalizable in an orthonormal basis (spectral theorem). That is, the eigenvectors are orthogonal (and can be normalized to 1). In other words, $$A.A^t=P^t.D.P$$
where $D$ is diagonal and $P$ verifies $$P^t.P=I$$
Here $I$ is the identity. So $P^t=P^-1$.

It is not automatic. The fact that a symmetric matrix $B$ is diagonalizable depends on two facts:

• all eigenvalues of $B$ are real




• the orthogonal of an eigenspace is invariant by $B$

Then you start with an eigenvalue and a one-dimensional eigenspace, and then the orthogonoal of said eigenspace is invariant, and now you can again find an eigenvalue and a one-dimensional eigenspace. The process continues until you exhaust the $n$ dimensions.

What the above achieves is that you obtain a basis of eigenvectors, which is precisely what diagonalization is. Actually, you can obtain an orthonormal basis of eigenvectors, which tells you that you can diagonalize $B$ using a unitary (i.e., an orthogonal matrix in this case).

Recall that a symmetric matrix is always diagonalizable by spectral theorem.

$(AB)^T=B^TA^T$. $therefore (AA^T)^T=(A^T)^TA^T=AA^T$

Thus $AA^T$ is symmetric. And symmetric matrices are diagonalizable.

symmetric matrices has $ n$ eigenvalues (not necessarily distinct) but how can I know if there is an invertible matrix $ P $ that diagonalize $ A $ ?

"$n$ eigenvalues (not necessarily distinct)" or, as I like to call it, "$n$ eigenvalues if counted with geometric multiplicity", means exactly that there is some basis of $Bbb R^n$ for which the linear map associated with $AA^t$ is diagonal: it's any linearly independent collection of $n$ eigenvectors, which the part "$n$ eigenvalues (not necessarily distinct)" tells us exists.