I've came across the following question:
Let $A \in M_2(\mathbb{R})$ and, for $X,Y \in M_{2\times 1}(\mathbb{R})$, define $f(X,Y) = Y^tAX$. Give sufficient and necessary conditions so that $f$ is a inner product.
I know that, since I have to show that $f$ defines an inner product over $\mathbb{R}$, $f(X,Y)$ must be bilinear, symmetric and positive definite. The bilinear part is pretty straightforward, but I don't know how to proceed with the problem. Any help would be appreciated.
Suppose $A$ is not symmetric, so that there exist $i$ and $j$ such that $a_{ij}\ne a_{ji}$. Can you construct $X$ and $Y$ such that $Y^tAX\ne X^tAY$?
Suppose $A$ is singular. Can you construct $X$ such that $X^tAX=0$?
Edited to add:
These two considerations show that $A$ has to be symmetric and non-singular. But that is not enough, because we might still have $X^tAX=0$ for some $X\ne 0$. We also need $A$ to be positive definite (as cmk points out in a comment).