if $\langle v , w \rangle = \sqrt {\langle v,v \rangle \langle w, w \rangle}$ then $v=kw$, $k \in \mathbb R$.

Question

Inspired by an earlier question.

It is clear to see that if $v=kw$ then $\langle v , w \rangle = \sqrt {\langle v,v \rangle \langle w, w \rangle}$, but why is the other direction also true?

This seems false, especially since I want to prove this for any inner product, not just the standard one.

$v,w \in \mathbb R^n$ are real, not complex.

Is it possible to prove this only with the properties of an inner product without actually defining it? Is it even true (for all inner products)? If not, is it at least true for the standard inner product? $\langle u,w \rangle = \sum_{i=1}^{n}v_i w_i$.

Answer 1

Any real inner product (that is, a bilinear map that is positive-definite: $\langle v,v \rangle > 0$ for any $v \neq 0$) satisfies the Cauchy-Schwarz inequality. Consider the following proof: $$ 0 \leqslant \langle au+bv,au+bv \rangle = a^2 \langle u,u \rangle + 2ab \langle u,v\rangle+b^2 \langle v,v \rangle, $$ with by definition equality when $au+bv=0$. We can choose $a=\sqrt{\langle v,v \rangle}$ and $b = \pm \sqrt{\langle u,u \rangle}$, and find $$ 0 \leqslant a^2 \langle u,u \rangle + 2ab \langle u,v\rangle+b^2 \langle v,v \rangle \\ = 2\sqrt{\langle v,v \rangle\langle u,u \rangle} \left( \sqrt{\langle v,v \rangle\langle u,u \rangle} \pm \langle u,v\rangle \right) $$ Hence $$ \sqrt{\langle v,v \rangle\langle u,u \rangle} \geqslant \langle u,v\rangle, \tag{1} $$ and the only inequality occurred when we said $0 \leqslant \langle au+bv,au+bv \rangle$. Hence equality occurs in (1) precisely when $au+bv=0$, which is when one of $\langle u,u \rangle$ and $\langle v,v \rangle $ is zero, i.e. one of $u,v$ is zero, or when $au+bv=0$, i.e. $u$ and $v$ are proportional.

Answer 2

Hint: Have a closer look at a common proof of the Cauchy–Schwarz inequality, which considers the minimum of the quadratic polynomial $\|v-xw\|^2=\langle v-xw,v-xw\rangle$. (Expand the polynomial and collect like powers of $x$ first.)

Answer 3

Suppose $\langle v, w \rangle=||v||. ||w||$, $w \neq 0$. We have to conclude that $v=kw$.

Take $c:=\frac{||v||}{||w||}$.

Then:

$$\langle v- cw, v-cw \rangle= ||v||^2-2c\langle v, w \rangle +|c|^2||w||^2$$ $$=2||v||^2-2c ||v||. ||w||=2 ||v||^2-2 ||v||^2=0.$$

Hence, $v-cw=0 \implies v=cw$. $\quad c$ is the $k$ you wanted.

OBS: Note that this is the natural candidate for $k$.