# 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.

• One approach: note that inner product (like any multilinear map) is determined by how it acts on a basis Commented Apr 25, 2015 at 23:06
• Combine with that the fact that every finite inner product space has an orthonormal basis. Commented Apr 25, 2015 at 23:13

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 $$u,u^{\prime} \in V$$, $$\langle u,u^{\prime} \rangle_{1}=c\langle u,u^{\prime} \rangle_{2}$$

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 $V$.