Prove that $\langle v,w \rangle = (A{\bf v}) \cdot (A{\bf w})$ defines an inner product on $\mathbb{R}^m$ iff $\ker(A)=\{{\bf 0}\}$

Question

My instructor showed the proof of this result by proving the three axioms of inner product on the given proposed inner product, then using positive definiteness to show the $\ker(A)=\{{\bf 0}\}$. I was wondering if someone could offer an alternative proof that proves both directions of the "if and only if" statement. That is, first assume that $\langle v,w \rangle = (A{\bf v}) \cdot (A{\bf w})$ defines an inner product, why does this mean the $\ker(A) = \{{\bf 0}\}$? Then assume $\ker(A) = \{{\bf 0}\}$, why does this mean $\langle v,w \rangle = (A{\bf v}) \cdot (A{\bf w})$ defines an inner product.

Answer 1

We have that

1. $$\begin{align}\langle\alpha v1+\beta v_2,w\rangle&=(A(\alpha v1+\beta v_2))\cdot(Aw)\\&=\alpha(Av_1)\cdot(Aw)+\beta(Av_2)\cdot(Aw)\\&=\alpha\langle v_1,w\rangle+\beta\langle v_2,w\rangle\end{align}$$

2. $$\langle v,w\rangle=(Av)\cdot(Aw)=(Aw)\cdot(Av)=\langle w,v\rangle$$

and

3. If $$0=\langle v,v\rangle=(Av)\cdot(Av)$$

we have $Av=0$. Since the kernel is zero we get $v=0$