example of a matrix A such that $\langle x,y \rangle$A$\neq \langle y,x \rangle$

Question

Let $A$ be a 3x3 matrix and x, y $\in R^3$ with x=($x_1,x_2,x_3$) and y=($y_1,y_2,y_3$). $\langle$x,y$\rangle$$A$=x($A$y), with x y the standard dotproduct.

I had to give an example of $A$ for which this was symmetrical, so $\langle$x,y$\rangle$$A$=$\langle$y,x$\rangle$$A$. The way the question was asked suggested that there are also matrices $A$ for which this isn't the case. I've tried to find such a matrix $A$, but can't think of any.

For what kind of $A$ is it not symmetrical?

Answer 1

A matrix $A$ is called symmetric if $A=A^T$. This is equivalent to

$x(Ay)=y(Ax)$ for all $x,y \in \mathbb R^3$.

Hence find a matrix $A$ with $A \ne A^T$.

Answer 2

In fact, you can show that $A=A^T$ (i.e. $A$ is symmetrical) if and only if $\forall x, y\in\Bbb R^3$ $$\langle x,y\rangle_A = \langle y,x\rangle_A$$