Which Matrix is an Inner Product We define the inner product of square matrices to be $\langle\vec x, \vec y\rangle_A=\vec x^TA\vec y$. One of the matrices $$\begin{bmatrix}
1 & 2 \\
2 & 1 \\
\end{bmatrix}$$ $$\begin{bmatrix}
2 & 1 \\
1 & 2 \\
\end{bmatrix}$$ violates the requirement $\langle\vec x, \vec x\rangle_A > 0$ for $\vec x \neq \vec 0$. I don't really understand the inner product in general, and I am really struggling to understand and prove which matrix isn't an inner product. Any help would be greatly appreciated.
 A: The first one does not generate an inner product because it has an eigenvalue of $-1$. You can check:
$$
        \begin{bmatrix}1 & -1 \end{bmatrix}
\begin{bmatrix}
1 & 2 \\
2 & 1 \\
\end{bmatrix}\begin{bmatrix} 1 \\ -1 \end{bmatrix} = -2
$$
The second one does generate an inner product.
A: An inner product is a way to map vectors to a scalair quantity. You can use it to 'multiply' vectors. To check wether this is an inner product, you'll have to go over the definition of an inner product and see if it suffices.
For the first matrix we have:
$\overrightarrow{x} = ({x_1, x_2})^T$ so : $< \overrightarrow{x}, \overrightarrow{y} >_A = (x_1 + 2x_2)y_1 + (2x_1 + x_2)y_2$
Now you need to check if this equals the complex conjugate of $< \overrightarrow{x}, \overrightarrow{y} >_A$, check linearity in the first argument, and check wether $0 \leq < \overrightarrow{x}, \overrightarrow{x} >_A$ and $ < \overrightarrow{x}, \overrightarrow{x} >_A = 0$ only if $\overrightarrow{x}$ is $\overrightarrow{0}$.
Same for the second matrix.
