Prove $\ AA^t $ is diagonalizable 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: 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). 
A: $A.A^t$ 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 $P^t=P^{-1}$.
A: $(AB)^T=B^TA^T$.  $\therefore (AA^T)^T=(A^T)^TA^T=AA^T$
Thus $AA^T$ is symmetric. And symmetric matrices are diagonalizable. 
A: Recall that a symmetric matrix is always diagonalizable by spectral theorem.
A: 
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.
