Given an $m \times n$ matrix $A$, it's easy to show that the matrix products $B = A^TA$ and $C = AA^T$ are both symmetric. I was wondering if any symmetric matrix with positive eigenvalues could be expressed as the product of some matrix and its transpose?
Note: It's obvious that if the eigenvalues are not all positive then this cannot be the case, as $A^TA$ and $AA^T$ both have to be positive semidefinite. However this is all I can think of regarding this problem.
Best Answer
Nevermind, I just got this. I'll leave my answer up in case anyone else finds it useful.
Any symmetric matrix $A$ has to be orthogonally diagonalisable, that is, $A = PDP^T$ for some diagonal matrix $D$ and orthogonal matrix $P$. In this case by the diagonalisation theorem, the diagonal values of $D$ are the eigenvalues of $A$, all of which are given to be $\geq 0$. Therefore, one can "square-root" $D$ as $D = M^2$, where $M$ is a diagonal matrix with the square roots of the diagonal values of $D$ on its diagonal, that is, $m_{ii} = \sqrt{d_{ii}}$ and $m_{ij} = 0$ if $j \neq i$. Therefore:
$$ A = PM^2P^T = (PM)(MP^T) = (PM)(M^TP^T) = (PM)(PM)^T $$