I'm reading in my Linear Algebra syllabus, and they write that given $A\in GL_n(\mathbb{R})$ and $A^T=A$ we can define the following inner product: $(y,z)_A=y^TAz$ for $y,z\in\mathbb{R}^n$.
I am a bit in doubt about this. Surely this map is linear in the first component and symmetric. However, I am in doubt about the map being positive-definite. Using the spectral theorem for symmetric matrices, we can find an orthonormal basis $\{v_1,\ldots,v_n\}$, and writing $x=\sum_{i=1}^nc_iv_i$ and writing out $(x,x)_A$ yields $\sum_{i=1}^nc_i^2\lambda_i$, but this still does not help for showing positiveness. Also, I tried rewriting to the standard inner product using the square root of $A$ but the problem here is that the resulting square root of the diagonal matrix might be complex if at least one of the eigenvalues is negative.
My question: is this really an inner product on $\mathbb{R}^n$? If so, how can I prove the map is positive-definite? I have the feeling we need that $A$ is positive semidefinite, but I'm not sure.
Of course it doesn't have to be positive-definite. Just let $A$ be the null matrix. Or let $A=-\operatorname{Id}$.