Let $A$ and $B$ be symmetric matrices. Prove:
- $AB=BA$
- $AB$ is a symmetric matrix
As for 1. due to the axiom $(AB)^T=B^T A^T$ so $AB=BA$
As for 2. I did not find any axiom that can support the claim, but from test I found that it is true for symmetric matrices when the entries on the diagonal are equal.
2 Answers
$\begingroup$Both claims are false and almost any $A$ and $B$ are counterexamples. For a specific example, you can see $$\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} \cdot \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 3 & 5 \end{pmatrix}$$ while $$\begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} \cdot \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 3 \\ 5 & 5 \end{pmatrix}.$$
$\endgroup$ 4 $\begingroup$Recall that by definition the product of two matrices (with components $A_{ij}$, $B_{ij}$) has components $(AB)_{ik}=\sum_j A_{i j} B_{jk}$. What about the components of ${BA}$? You'll also need to see what being symmetric implies about the components $A_{ij}$.
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