Multiplication of two symmetric matrices may not be symmetric

Let say we have two symmetric matrices $$A, B \in \mathbf{R}^{n \times n} \\ A=A^T, B=B^T$$ show that $AB$ may not be symmetric?

my solution:

furthermore we know that every symmetric matrix $X$ can be written as: $$X = \frac{1}{2} (Y + Y^T),\text{where}\ Y \in \mathbf{R}^{n \times n}$$

let say we have to arbitary squre matrices $$M, W \in \mathbf{R}^{n \times n}$$ So we can express $A, B$ as: $$ A = \frac{1}{2} (M + M^T) \\ B = \frac{1}{2} (W + W^T) $$ So the product of $A,B$ will be: $$ C=AB = \frac{1}{2} (M^T + M) \frac{1}{2} (W^T + W) \\ = \frac{1}{4} (MW + M W^T + M^T W + M^T W^T) \in \mathbf{R}^{n \times n} $$

For $AB$ to be symmetric we must show that $AB=(AB)^T=B^T A^T$

$$ C^T=(AB)^T = B^T A^T= \frac{1}{2} (W^T + W) \frac{1}{2} (M^T + M) \\ = \frac{1}{4} (W^T M^T + W^T M + W M^T + WM) \in \mathbf{R}^{n \times n} $$

And since matrix multiplication is not commutative $C$ is not necessary equal to $C^T$.

Is that the right way to proof it?

Whta are other proof concepts?

thanks

The simplest way to prove it is to provide an example:

Let $A = \begin{pmatrix}1&0\\0&0\end{pmatrix},~ B = \begin{pmatrix}1&1\\1&1\end{pmatrix}$. Then $A$ and $B$ are symmetric, but not $A\cdot B = \begin{pmatrix}1&1\\0&0\end{pmatrix}$.

Your approach ("for some symmetric matrices $A, B$ the product $A\cdot B$ can be asymmetric") is not enough, since it does not actually imply anything at all. You need to show "for some symmetric matrices $A, B$ the product $A\cdot B$ is asymmetric". Of course, a single example choice of $A$ and $B$ is enough to prove the statement, so it's usually not worth the effort to construct a whole class of examples.

If you like the idea of seeing why the statement does not hold in general, here is a simpler way to construct infinitely many counterexamples:

Let $A$ and $B$ be such that $A^T = A$ and $B^T = B$. Then $(AB)^T = B^TA^T = BA$, so $AB$ is symmetric if and only if $AB = BA$. Then choose any non-commutative $A$ and $B$; since $AB \neq BA$ we know that $(AB)^T \neq AB$.