Can a product of symmetric matrices give a non-zero skew symmetric matrix? I've been trying to find symmetric $\mathbf{A},\mathbf{B}$ such that $\mathbf{AB}$ is skew-symmetric, but it seems that no matter what I try, I end up forcing $\mathbf{AB}=\mathbf{0}$. Is it possible for this product to be non-zero?
I've also tried proving that $\mathbf{AB}$ must be $\mathbf{0}$ but haven't got much further than $\mathbf{AB}=-\mathbf{BA}$.
 A: Let
$$
A=\begin{bmatrix}1&0\\0&-1\end{bmatrix},\ \ \ B=\begin{bmatrix}0&1\\1&0\end{bmatrix}.
$$
Then
$$
BA=-AB=\begin{bmatrix}0&-1\\1&0\end{bmatrix}.
$$
A: There is a beautiful theorem in Kaplansky's book - Linear Algebra and Geometry: A Second Course - that answers your question.
Thm 66: Let A be a square matrix over an arbitrary field. Then A is symmetrically similar to its transpose $A^t$, and A can be expressed as a product of two symmetric matrices. 
See page 76.
Every matrix is  a product of two symmetric matrices!
A: One can easily characterise all pairs of real symmetric matrices $(A,B)$ such that the product $AB$ is skew-symmetric. By an orthogonal change of basis, we may assume that $A$ is of the form
$$
A=\lambda_1 I_{m_1} \oplus \lambda_2 I_{m_2} \oplus \cdots \oplus \lambda_k I_{m_k} \oplus 0_{m_{k+1}\times m_{k+1}},
$$
where


*

*$\lambda_1,\ldots,\lambda_k$ are distinct, nonzero and real,

*for some $2s\le k$, we have $\lambda_{j+1}=-\lambda_j$ for each odd index $j\le 2s$, and

*when $2s<j\le k$, the number $-\lambda_j$ is not an eigenvalue of $A$.


(In other words, on the diagonal of $A$, pairs of blocks with positive and negative $\lambda$s of the same moduli appear before the "orphan" blocks whose eigenvalues have no negative counterparts).
When an $n\times n$ symmetric matrix $A$ is written in such a normal form and $B$ is partitioned into a block form accordingly, it is easy to see that $AB$ is skew symmetric if and only if $B$ has the form
$$
B=\left[\begin{array}{cc|c|cc|c|c}
0_{m_1\times m_1}&B_1\\
B_1^T&0_{m_2\times m_2}\\
\hline
&&\ddots\\
\hline
&&&0_{m_{2s-1}\times m_{2s-1}}&B_s\\
&&&B_s^T&0_{m_{2s}\times m_{2s}}\\
\hline
&&&&&\Large 0_{m\times m}\\
\hline
&&&&&&\Large S
\end{array}\right]
$$
where each $B_j\ (j=1,2,\ldots,s)$ is an arbitrary $m_{2j-1}\times m_{2j}$ real matrix, $S$ is any real symmetric matrix of size $0_{m_{k+1}\times m_{k+1}}$ and the large zero square subblock preceding $S$ is of size $m=\sum_{j=2s+1}^k m_j$.
