Suppose $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$ are inner products on $V$ such that... Suppose $\langle\cdot,\cdot\rangle_1$ and $\langle\cdot,\cdot\rangle_2$ are inner products on $V$ such that $\langle v,w\rangle_1=0$ if and only if $\langle v,w\rangle_2=0$. Prove that there is a positive number $c$ such that $\langle\cdot,\cdot\rangle_1=c\langle\cdot,\cdot\rangle_2$ for every $v,w\in V$.
I'm at a loss on how to start this. Any guidance is appreciated.
 A: If $V=\{0\}$ then there is nothing to prove. Suppose that $V \neq \{0\}$ and let $v \in V$ with $v \neq 0$. Then both $\langle v,v \rangle_{1}$ and $\langle v,v \rangle_{2}$ are nonzero. Let $$c = \frac{\langle v,v \rangle_{1}}{\langle v,v \rangle_{2}}$$ Now let $w \in V$ with $\langle w,v \rangle_{1} \neq 0$ (if $\langle w,v \rangle_{1} = 0$ then there is nothing to prove). By the orthogonal decomposition (Axler (2015) p. 171), $\langle w-\frac{\langle w,v \rangle_{2}}{\langle v,v \rangle_{2}}v,v \rangle_{1} =\langle w-\frac{\langle w,v \rangle_{2}}{\langle v,v \rangle_{2}}v,v \rangle_{2} =0$, and we obtain $$\frac{\langle w,v \rangle_{1}}{\langle w,v \rangle_{2}} = \frac{\langle v,v \rangle_{1}}{\langle v,v \rangle_{2}}$$ Similarly, since $\langle w,v-\frac{\langle v,w \rangle_{2}}{\langle w,w \rangle_{2}}w \rangle_{1} =\langle w,v-\frac{\langle v,w \rangle_{2}}{\langle w,w \rangle_{2}}w \rangle_{2} =0$, we obtain $$\frac{\langle w,v \rangle_{1}}{\langle w,v \rangle_{2}} = \frac{\langle w,w \rangle_{1}}{\langle w,w \rangle_{2}}$$ That is, for any $w \in V$, $$\frac{\langle w,w \rangle_{1}}{\langle w,w \rangle_{2}}=\frac{\langle v,v \rangle_{1}}{\langle v,v \rangle_{2}}=c$$ Here we have shown that for any $u,u^{\prime} \in V$, $$\langle u,u^{\prime} \rangle_{1}=c\langle u,u^{\prime} \rangle_{2}$$
A: first part:
Assume that $$
\langle x, y \rangle_1 = \langle x, y \rangle_2 = 0 \\
\langle x, x \rangle_2 = \langle y, y \rangle_2 = 1
\langle x, x \rangle_1 = c \\
\langle y, y \rangle_1 = d \\
$$
Then:
$$
\langle x + y, x - y \rangle_2 = 0
\\ \implies 0 = \langle x + y, x - y \rangle_1
= \langle x, x \rangle_1 - \langle y, y \rangle_1 
$$
second part:
consider a finite dimensioned subspace $V'$, an 
$\langle ., . \rangle_1$-orthonormal basis 
$(e_1\dots e_d)$ if $V'$.
It is also a $\langle ., . \rangle_2$-orthogonal basis;
its matrix is, according to the first part, $ a I_d$, for some $a>0$; or:
$$
 \langle x, y \rangle_1 = a \langle x, y \rangle_2 
$$
It is true for any $V'$, hence it remains true on $V$.
