TITLE: How to express a matrix as a product of two symmetric matrices?
QUESTION [6 upvotes]: Let $A$ be a matrix and $J$ its Jordan canonical form. How can one express $A$ as a product of two symmetric matrices?
I expressed $J$ as a product of two symmetric matrices: block by block in the following manner:
$$\begin{bmatrix}\lambda&1 & \\ &\lambda&1 \\ & &\lambda \end{bmatrix} =\begin{bmatrix} & & 1\\ &1& \\1& & \end{bmatrix}\begin{bmatrix} & & \lambda\\ &\lambda&1 \\\lambda& 1& \end{bmatrix}$$
and then I was trying to make use of the fact that $A$ and $A^t$ are similar. Yet I got stuck.
Any hints are hugely appreciated.

REPLY [5 votes]: Apparently, the tricks of yours and Igor's can be extended to the real case, because a real square matrix $A$ is $\mathbb R$-similar to its real Jordan form, whose Jordan blocks for conjugate pairs of nonreal eigenvalues $a\pm ib$ are block-triangular matrices of the form
$$
\begin{bmatrix}C_2&I_2\\ &C_2&I_2\\ &&\ddots&\ddots\\ &&&C_2&I_2\\ &&&&C_2\end{bmatrix}
=\begin{bmatrix}&&&&D_2\\ &&&D_2\\ &&⋰\\ &⋰\\ D_2\end{bmatrix}
\begin{bmatrix}&&&&H_2\\ &&&H_2&D_2\\ &&⋰&⋰\\ &⋰&⋰\\ H_2&D_2\end{bmatrix}
$$
where $C_2=\begin{bmatrix}a&-b\\ b&a\end{bmatrix}$, $\ I_2$ is the identity matrix, $D_2=\begin{bmatrix}1\\ &-1\end{bmatrix}$ and $H_2=\begin{bmatrix}a&-b\\ -b&-a\end{bmatrix}$.