Let $\langle \cdot, \cdot \rangle_{1}$ and $\langle \cdot,\cdot \rangle_{2}$ be inner product on a finite-dimensional vector space with the property that \begin{align*} \langle u,v\rangle_{1}=0 \Leftrightarrow \langle u,v\rangle_{2}=0 \end{align*} for all $u,v\in V$. Show that there is a scalar $a\in F$ for which \begin{align*} \langle u,v\rangle_{2}=a\langle u,v\rangle_{1} \end{align*}

I saw a solution wrote like this:

Presumably, $F=\mathbb {R}$ or $\mathbb {C}$. For every nonzero vector $v\in V$, define $c_{v}=\frac {\langle v,v\rangle_{2}}{\langle v,v\rangle_{1}}$. Now, for any $v,w\in V$, let $x=w-\frac {\langle w,v\rangle_{1}}{\langle v,v\rangle_{1}}v$. Then $\langle x,v\rangle_{1}=0$. Hence, $\langle x,v\rangle_{2}=0$ and in turn \begin{align*} \langle w,v\rangle_{2}&=\frac {\langle w,v\rangle_{1}}{\langle v,v\rangle_{1}}\langle v,v\rangle_{2}\\ &=\frac {\langle v,v\rangle_{2}}{\langle v,v\rangle_{1}}\langle w,v\rangle_{1}\\ &=c_{v}\langle w,v\rangle_{1} \end{align*} for all nonzero $v,w\in V$. Therefore, \begin{align*} c_{v}\langle w,v\rangle_{1}&=\langle w,v\rangle_{2}\\ &=\overline {\langle v,w\rangle_{2}}\\ &=\overline {c_{w}\langle v,w\rangle_{1}}\\ &=c_{w}\langle w,v\rangle_{1} \end{align*} for all $v,w\not=0$. Now, for any $v,w\not =0$, there exists some $y\in \{w+tv:t\in \mathbb {R}\}$ such that $\langle w,y\rangle_{1}$, $\langle y,v\rangle_{1}\not =0$. Hence, we have $c_{v}\langle y,v\rangle_{1}=c_{y}\langle y,v\rangle_{1}$ and $c_{y}\langle w,y\rangle_{1}=c_{w}\langle w,y\rangle_{1}$. Hence $c_{v}=c_{y}=c_{w}$ for all $v,w\not=0$. In other word, all the $c_{v}$'s are equal to some common constant $c>0$. Thus, the proof is done. But I don't know what if $\langle w,v\rangle_{1}=0$ and why do we care bout the vectors in $\{w+tv:t\in \mathbb {R}\}$.

  • $\begingroup$ If $\langle w,v\rangle_1=0$, then $\langle w,v\rangle_2=0$ and $0=c\times 0$ holds for any $c$, so I don't see any problem in that case. $\endgroup$ – learner Mar 16 at 5:53
  • $\begingroup$ @learner Then there's no way to find the value of c. $\endgroup$ – Jiexiong687691 Mar 16 at 6:18
  • 1
    $\begingroup$ You don't really need to find a particular value of $c$, you have to show that there exists at least one such value that works. In the above case, any value of $c$ works, so we're done. $\endgroup$ – learner Mar 16 at 6:32

What if $\langle w,v\rangle_1 = 0$?

The idea is this : if $\langle w,v\rangle \neq 0$ (and we know for sure that such $w$ exists, simply taking $w = v$), then we can get $c_v$ from there. Now, does that $c_v$ work when $\langle w,v\rangle_1 = 0$, is the question.

But, if $\langle w,v \rangle_1 = 0$ then $\langle w,v\rangle_2 = 0$ by the given conditions, so in fact $\langle w,v\rangle_1 = c_v \langle w,v\rangle_2$ (obviously because $0 \times c_v = 0$).

In short, it will still hold that for all non-zero $v$ and $w$ that $\langle w,v\rangle_2 = c_v \langle w,v\rangle_1$,because it will hold through the given argument if $\langle w,v\rangle_1 \neq 0$, and both sides are zero if $\langle w,v \rangle_1 = 0$.

Why do we care about the vectors $w+tv$?

The best way to put it is this : if we show for any $v,w$ that $c_v = c_w$, then we are done. If $w=cv$ for some scalar $c$ then we can easily show that $c_w = c_v$.

If $w$ and $v$ are linearly independent, then consider the two dimensional subspace generated by $v$ and $w$, and restrict the inner products $1$ and $2$ to this space. The point is , that the conditions for the inner products still hold in this two dimensional subspace. So the conclusion must hold for this subspace. Therefore, using only the vectors in this two dimensional subspace, we should be able to conclude that $c_v = c_w$. So we care about these vectors, because the proof should require only the properties of these vectors to conclude.

Which is why we restrict our attention to the vectors $\{w+tv : t \in \mathbb R\}$, which (up to scaling) covers every element of the span of $v$ and $w$. Then the rest should be straightforward.


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