1
$\begingroup$

Let $A \in \mathbb{R}^{n \times n}$ be symmetric. I am trying to understand under which conditions on $B \in \mathbb{R}^{n \times n}$ 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?

$\endgroup$
2
$\begingroup$

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 $$ C_{11}\quad \text{is}\quad \left\{ \begin{array}{l} \text{an arbitrary matrix if}\quad \alpha_1=0\\ \text{an arbitrary symmetric matrix if}\quad \alpha_1\ne 0. \end{array} \right. $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.