I'm working through Axler's "Linear Algebra Done Right", but I'm stuck on Exercise 6.B Q11:
Suppose $<.,.>_1$ and $<.,.>_2$ are inner products on $V$ such that $<u,v>_1 = 0 \iff <u,v>_2 = 0$ Prove that $<v,w>_1 = c<v,w>_2$ for some fixed $c>0$ for any vectors v and w.
I've been able to simplify the problem as so:
Let $k$ be a scalar such that $u-kv$ is orthogonal to $v$, so we have
$$ \frac{<u,v>_1}{<u,v>_2} = \frac{<kv,v>_1 + <u-kv,v>_1}{<kv,v>_2 + <u-kv,v>_2}$$ $$ = \frac{<v,v>_1}{<v,v>_2}$$
So it would suffice to show that $||v||_1 = c||v||_2$. However I'm stuck on how to show this.
Thanks to Gram-Schmidt you can take a basis $\{v_1,...,v_n\}$ which is orthonormal with respect to the first inner product. Now, thanks to orthonormality, you have: $$<v_i+v_j,v_i-v_j>_1=<v_i,v_i>_1-<v_j,v_j>_1=1-1=0$$ Since $<.,.>_2$ has the same orthogonal vectors $v_i+v_j,v_i-v_j$ are also orthogonal w.r.t $<.,.>_2$. So: $$0=<v_i+v_j,v_i-v_j>_2=<v_i,v_i>_2-<v_j,v_j>_2 \Rightarrow <v_i,v_i>_2=<v_j,v_j>_2=c$$ So, $\{\frac{v_1}{c},...,\frac{v_n}{c}\}$ is an orthonormal basis w.r.t the second inner product and, by linearity, for every vectors $v$ and $w$ you have $<v,w>_1=c<v,w>_2$