Product of symmetric matrices Let $A \in \Bbb{R}^{n \times n}$ be symmetric. I am trying to understand under which conditions on $B \in \Bbb{R}^{n \times n}$ the product $AB$ is also symmetric. It is clear that if $B$ is symmetric and commutes with $A$, we have
$$
(AB)^T = B^TA^T = BA = AB.
$$
Do you see whether the result still holds under weaker conditions?
 A: Let $A\in {\mathbb R}^{n\times n}$ a symmetric matrix. Then there exists an orthogonal matrix $Q\in {\mathbb R}^{n\times n}$ such that $A=QDQ^T$ where $D\in {\mathbb R}^{n\times n}$ is
a diagonal matrix. Moreover, we may assume that $D=\alpha_1 I_{n_1}\oplus \cdots \alpha_n I_{n_k}$, where $\alpha_i\in {\mathbb R}$ $(i=1, \ldots, k)$ are distinct eigenvalues of $A$
and $n_1+\cdots+n_k=n$.
Assume that $B\in {\mathbb R}^{n\times n}$ is such that $AB$ is symmetric. Then
$$ AB=(AB)^T=B^TA^T=B^T A. $$
Hence $B$ has to satisfy the condition
$$ AB=B^TA \tag1.$$
It is obvious that the converse holds as well: if $B\in {\mathbb R}^{n\times n}$ satisfies (1), then $AB$ is symmetric. Note that it follows thata symmetric $B\in {\mathbb R}^{n\times n}$  commutes with $A$ if and only if $AB$ is symmetric. Let us look for non-symmetric $B\in {\mathbb R}^{n\times n}$ satisfying (1).
If we replace $A$ by $QDQ^T$, then (1) read as
$$ QDQ^TB=B^TQDQ^T$$
which gives
$$ D(Q^TBQ)=Q^TB^TQD=(Q^TBQ)^TD. $$
It follows that $C\in {\mathbb R}^{n\times n}$ satisfies condition
$$ DC=C^TD \tag 2$$
if and only if $B=QCQ^T$ satisfies (1). Hence it is enough to consider (2). Write $D$ as a
block diagonal matrix
$$ D=\left( \begin{array}{cccc}
\alpha_1 I_{n_1} & 0 & \ldots & 0\\
0 &\alpha_2 I_{n_2} & \ldots & 0\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \ldots & \alpha_k I_{n_k}
\end{array} \right) $$
and assume that
$$ C=\left( \begin{array}{cccc}
C_{11} & C_{12} & \ldots & C_{1k}\\
C_{21} &C_{22} & \ldots & C_{2k}\\
\vdots & \vdots & \ddots & \vdots\\
C_{k1} & C_{k2} & \ldots & C_{kk}
\end{array} \right) $$
satisfies (2). Then one has
$$ \alpha_i C_{ij}=\alpha_j C_{ji}^{T}. \tag3$$ 
Let us agree that $\alpha_1=0$ if $0\in \{ \alpha_i;\, 1\leq i\leq k\}$. Then
$$ C=\left( \begin{array}{cccc}
C_{11} & C_{12} & \ldots & C_{1k}\\
\frac{\alpha_1}{\alpha_2}C_{12} &C_{22} & \ldots & C_{2k}\\
\vdots & \vdots & \ddots & \vdots\\
\frac{\alpha_1}{\alpha_k}C_{1k} & \frac{\alpha_2}{\alpha_k}C_{2k} & \ldots & C_{kk}
\end{array} \right) $$
where $C_{ij}$ $(1\leq i<j\leq k)$ are arbitrary matrices, $C_{22}, \ldots, C_{kk}$
atre arbitrary symmetric matrices and 
$$ C_{11}\quad \text{is}\quad \left\{ \begin{array}{l} \text{an arbitrary matrix if}\quad \alpha_1=0\\
 \text{an arbitrary symmetric matrix if}\quad \alpha_1\ne 0. \end{array} \right. $$
