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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.

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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.

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  • $\begingroup$ Thank you for the reply! Does the matrix need to be positive semi-definite (=non-negative eigenvalues) because if we select $x=y$ there may be a violation of the inner product's definition? math.stackexchange.com/questions/518890/… $\endgroup$ – Konstantin Feb 16 '18 at 6:53
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    $\begingroup$ @Konstantin : The matrix must be positive definite in order for $[x,y]=x^{T}Ay$ to be an inner product. Positive semi-definite would mean that $[x,x]=0$ could hold without $x=0$, even though all other required properties of an inner product would be satisfied. $\endgroup$ – DisintegratingByParts Feb 16 '18 at 7:13
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    $\begingroup$ Thank you very much for your explanation, my mistake! $\endgroup$ – Konstantin Feb 16 '18 at 7:18
  • $\begingroup$ A couple more questions if possible: any positive definite square matrix represent an inner product or there are some additional requirements? Would proper matrices of same dimensions define essentially the same inner product (norms in finite dim. spaces are the same up to isomorphism)? $\endgroup$ – Konstantin Feb 16 '18 at 7:22
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    $\begingroup$ @Konstantin : Any positive definite square matrix defines an inner product. All norms are equivalent on a given finite-dimensional space, meaning that there are positive constants $A$, $B$ such that $\|x\|_1 \le A\|x\|_2$ for all $x$ and $\|x\|_2 \le B\|x\|_1$ for all $x$. $\endgroup$ – DisintegratingByParts Feb 16 '18 at 8:36
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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.

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