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In this answer I proved that for $V$ a vector space over a field $F$ of finite dimension and $\langle ,\rangle _1$, $\langle ,\rangle _2$ two inner products defined on it such that

$$\langle v,w\rangle _1=0\iff \langle v ,w\rangle _2=0 \tag{H}.$$

Then $\langle v,w\rangle _1=c\langle v,w\rangle _2$ for some scalar $c$.

Is the result still holding for infinite dimensional vector spaces?

mathcounterexamples.net
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    The result for infinite-dimensional spaces follows. A counterexample would be $<a,b>_1=c_1<a,b>_2$, $<c,d>_1=c_2<c,d>_2,$ $c_1\ne c_2$. But $a,b,c,d$ are contained in some finite-dimensional subspace – David C. Ullrich May 21 '18 at 20:32

1 Answers1

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I believe the answer is yes.

Lemma:

Let $V$ be a vector space and $f,g$ two linear functionals on $X$ such that $\ker g \subseteq \ker f$. Then there exists a scalar $c$ such that $f = cg.$

Now, fix $w \in V$ and consider the linear functionals $v \mapsto \langle v, w\rangle_1$ and $v \mapsto \langle v, w\rangle_2$. Their kernels are equal so there exists a scalar $c(w)$ depending on $w$ such that $$\langle v, w\rangle_1 = c(w) \langle v, w\rangle_2, \forall v \in V$$

Now, using symmetry:

$$c(w) \langle v, w\rangle_2 = \langle v, w\rangle_1 = \overline{\langle w, v\rangle_1} = \overline{c(w) \langle w, v\rangle_2} = \overline{c(v)}\langle v, w\rangle_2$$

We conclude $c(w) = \overline{c(v)}$ for all $w,v$ such that $\langle v, w\rangle \ne 0$. Since we can vary $v$ and $w$ independently, we conclude that $c(w)$ is actually a constant scalar $c$.

Hence $\langle v, w\rangle_1 = c \langle v, w\rangle_2$ for all $v,w \in V$.

mechanodroid
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