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I want to prove the following which seems reasonable to me:

$\mathbf{A}v\cdot \mathbf{A}w = 0 \iff v \cdot w =0$ given $\det(A) \ne 0$

If determinant is zero it's easy, $\mathbf{0}$ matrix falsifies the claim. The best I could come up with is:

$\mathbf{A}v\cdot \mathbf{A}w = v^\intercal \mathbf{A}^\intercal\mathbf{A}w$ and since $\mathbf{A}^\intercal\mathbf{A}$ is symmetric and full rank it's diagonalizable thus I can write it as $v^\intercal \mathbf{P}^{-1}\Lambda\mathbf{P}w$ and I have a feeling this could be useful somehow, but I lack further knowledge how to proceed.

P.S.

Linear algebra isn't my strong suit...so it might appear trivial to you, though I'm still learning mostly on my own

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    Are you sure, that it is correct? Consider $v = (1,1), w=(1, -1)$ and $A$ a diagonal matrix that isn't a multiple of the identity. – 2017-02-01
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    This is false. Clearly $v=\begin{bmatrix} 1\\ 0\end{bmatrix}$ and $w=\begin{bmatrix} 0\\ 1\end{bmatrix}$ are orthogonal, but not after multiplying them by $A=\begin{bmatrix} 1 & 1\\ 1 & 2\end{bmatrix}$. In order to satisfy your condition, the matrix should be unitary (i.e., $A^*A=I$). – 2017-02-01

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You've made an interesting guess, but your statement is false. For example: take $$ A = \pmatrix{1&0\\0&2} $$ clearly, $\det(A) = 2 \neq 0$. However, when we take $v = (1,1)$ and $w = (1,-1)$, we find that $v \cdot w = 0$, but $Av \cdot Aw \neq 0$.

The correct statement is as follows:

($Av \cdot Aw = 0 \iff v \cdot w = 0$) holds if and only if $A$ is a multiple of an orthogonal matrix. That is, it holds if and only if $A^TA$ is a multiple of the identity matrix.