Square matrices as products of symmetric matrices

Let $\mathbb{F}$ be any field and let $A$ be a square matrix over $\mathbb{F}$. Then $A$ is the product of two symmetric matrices over $\mathbb{F}$.

Solution:

 
(We give the proof as given by Heydar Radjavi in American Mathematical Society Journal) 

If $S_1, S_2, T$ are matrices of the same size over $\mathbb{F}$, if $S_j = S_j^{t}, \; j=1, 2, \dots, n $ and if $T$ is invertible then $T^{-1} (S_1 S_2) T$ is the product of two symmetric matrices $T^{-1} S_1 (T^t)^{-1}$ and $ T^t S_2 T$ over $\mathbb{F}$. Hence we can assume , without loss of generality, that $A$ is in the rational form :

$$A= {\rm diag} (A_1, \cdots A_n )$$

where each $A_i$ is of the form:

$$ B=\begin{pmatrix}
0 &0  &0  &\cdots  &0  &a_1 \\
 1& 0 &0  &\cdots  &0  &a_2 \\
 0&1  &0  &\cdots  &0  &a_3 \\
 \vdots& \vdots & \vdots &\ddots  &\vdots  &\vdots \\
 0& 0 &0  &\cdots  &1  &a_n
\end{pmatrix}$$

So it suffices to prove the theorem for a matrix of the form $B$.  The matrix:

$$C=\begin{pmatrix}
a_2 &a_3  &a_4  &\cdots  &a_{n-1}  &a_n  & -1\\
 a_3&a_4  &a_5  &\cdots  &a_n  &-1  &0 \\
 \vdots&\vdots  &\vdots  & \ddots &\vdots  &\vdots  &\vdots \\
 a_n &-1  &0  &\cdots  &0  &0  &0 \\
 -1&0  &0  &\cdots  &0  &0  &0
\end{pmatrix}$$

is symmetric and invertible over $\mathbb{F}$. Computation shows that $BC$ is also symmetric and $B= (BC)C^{-1} $ is the product of two symmetric matrices over $\mathbb{F}$.