How to express a matrix as a product of two symmetric matrices?

Question

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.

Answer 1

Writing your matrices as $S, C$ and your original matrix is $A,$ note that

$$A = B J B^{-1} = BSCB^{-1} = BSB^t(B^t)^{-1} C B^{-1} = [BSB^t] [(B^{-1})^t C B^{-1}].$$ This works for COMPLEX matrices (since you need the complex numbers for the Jordan canonical form). The fact is true for real symmetric matrices, however, and you should look at the paper by Bosch (from which I stole the previous argument, too).

Answer 2

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 $H_2=\begin{bmatrix}a&-b\\ -b&-a\end{bmatrix}$.