Let $V$ be a vector space and let$ \langle,\rangle_1$ and $\ \langle,\rangle_2$ be inner products in $V$ s.t. $\langle,\rangle_1=0 \iff \langle,\rangle_2=0$.
Prove $ \langle v,w\rangle_1=c\langle v,w\rangle_2$ for every $v,w \in V$.
I've been struggling coming up with a solution for this. This is what I have so far:
Let $U$ be a subspace of $V$. We know $\ U \oplus U^{\bot}=V$ for $\langle,\rangle_1$ and $\langle,\rangle_2$.
Let $\ B=\{u_1,...u_n\}$ be an orthonormal basis for $U$ and $\ C=\{v_1,...v_k\}$ an orthonormal basis for $\ U^{\bot}$ under $\langle,\rangle_1$.
$\ B'=\{\frac{u_1}{||u_1||},...,\frac{u_n}{||u_n||}\}$ is an orthonormal basis for $U$ and $\ C'=\{\frac{v_1}{||v_1||},...,\frac{v_k}{||v_k||}\}$ an orthonormal basis for $\ U^{\bot}$ under $\langle,\rangle_2$.
Take$\ u \in B $ and $\ v \in C$. If I prove for base vectors the argument is valid for any vector in the subspace.
I'd like to find the ratio using the norms of the bases B' and C', I'm not sure how to get at it and if this is the right direction.
$\endgroup$ 23 Answers
$\begingroup$You are going in the right direction, but there are some unnecessary complications.
There is no need to split the space into a direct sum. Just pick an orthonormal basis $\{v_1,v_2,\ldots,v_N\}$ of $V$ (suppose its dimension is $N$) with respect to $\langle,\rangle_1$. Then this basis is at least an orthogonal basis with respect to $\langle,\rangle_2$.
Now, for every pair of distinct $i,j$, consider $x=v_i+v_j$ and $y=v_i-v_j$ and their inner products with respect to both $\langle,\rangle_1$ and $\langle,\rangle_2$. The rest should be straightforward.
$\endgroup$ $\begingroup$Assuming that $V$ is a vector space over $\Bbb R$. Let $v(\neq 0)\in V$
Define $f_1:V\to \Bbb R$ by $f_1(w)= \langle v,w\rangle_1$
Also Define $f_2:V\to \Bbb R$ by $f_2(w)= \langle v,w\rangle_2$
Surely $f_1,f_2\neq 0$
Now $\langle v,w\rangle_1=0\iff \langle v,w\rangle_2\implies \ker f_1=\ker f_2$ .
Since $f_1\neq 0$ so there exists $x_0\in V$ such that $f(x_0)\neq 0$
Let $x\in V$ then any $x=\alpha x_0+y;y\in \ker f_1 ,\alpha=\dfrac{f_1(x)}{f_1(x_0)}$.
Since $\ker f_1=\ker f_2\implies y\in \ker f_2$
So $f_2(x)=\alpha f_2(x_0)=\dfrac{f_1(x)}{f_1(x_0)}f_2(x_0)=\beta f_1(x)$.
And hence $\langle v_1,v_2\rangle_1=\beta \langle v_1,v_2\rangle_2\text{Q.E.D}$
$\endgroup$ $\begingroup$You do not need to assume finite dimensionality of $V$. The result you are trying to prove holds in any inner product space, real or complex, of any (possibly infinite) dimension. Summoning an orthonormal basis just complicates the problem. You should start by figuring out what the number $c$ must be. The answer is $c=\left\langle{u,u}\right\rangle_1/\left\langle{u,u}\right\rangle_2$, where $u$ is any nonzero vector in $V$ (If $V$ is the zero space, the statement to be proved is trivially true, so we can assume $V$ has nonzero vectors). For this definition of $c$ to make sense however, you have to show that it does not depend on the choice of $u$. In other words, you have to show that $$\frac{\left\langle{u,u}\right\rangle_1}{\left\langle{u,u}\right\rangle_2}=\frac{\left\langle{v,v}\right\rangle_1}{\left\langle{v,v}\right\rangle_2}$$ holds for all nonzero $u,v\in V$. Once you've done that, showing that $c$ is the desired ratio is straightforward.
$\endgroup$
