Let $v,u\in\mathbb{R}^n$ and $A\in M_n\left(\mathbb{R}\right)$ an invertible matrix.
Prove $\left\langle v | u \right\rangle = v\cdot (A^tAu) $ is an inner product over the reals.
I was able to (allegedly) prove both symmetry, linearity and positive-definiteness without using the invertibility of $A$ at all... Linearity is straightforward; symmetry I showed as follows: $$\left\langle v | u \right\rangle = v\cdot (A^tAu) = (A^tAu)\cdot v = (A^tAu)^t v = (u^tA^t(A^t)^t)v$$ $$ = u^t(A^tAv) = u\cdot(A^tAv) = \left\langle u | v \right\rangle$$ And finally for positive-definiteness I identified $A^tA$ as a symmetric matrix $M$ with non-negative entries in the main diagonal: $$ A^tA \equiv M \begin{bmatrix}a^2_{11} & a_{12} & \dots & a_{1n}\\a_{12} & a^2_{22} & \dots & a_{1n}\\ \vdots & & \ddots & \vdots\\a_{1n}& \dots & & a^2_{nn}\end{bmatrix} $$ yielding in short $$\left\langle v | v \right\rangle = \sum_{i=1}^na_{ii}^2v_i^2 + 2\sum_{i=1}^n\sum_{j\neq i}a_{ij}v_iv_j = \left(\sum_{i,j=1}^na_{ij}v_iv_j\right)^2\geq0$$ Is there anything I'm missing in my proof?
Edit: I think I got it - A needs to be invertible for $\left\langle v | u \right\rangle = 0 $ to hold iff $v=0$, otherwise it can be the zero matrix. Is there another requirement for the invertibility?
You need to show that if $v$ is not zero then $\langle v, v\rangle > 0$.
This requires the assumption that the null space of $A $ is trivial.
(Note that $ \langle v,v \rangle = v^T A^T Av = (Av)^T Av = \| Av \|_2^2$.)