Show that $u$ is a multiple of $v$ or vice versa in the following

Question

Show that if $\left | \left \langle u,v \right \rangle \right | = \left\|u \right \| \left\|v \right \|$ then either $u$ is a multiple of $v$ or $v$ is a multiple of $u$.

Here is what I did:

Assume $u\neq0\neq v$ (else the result is trivially true)

$\left | \left \langle u,v \right \rangle \right | = \left\|u \right \| \left\|v \right \|$ $\iff \left \langle u,v \right \rangle . \overline{\left \langle u,v \right \rangle} = \left \langle u,u \right \rangle . \left \langle v,v \right \rangle$

Since $u\neq0\neq v$, thus

$\left \langle u,v \right \rangle . \frac{\overline{\left \langle u,v \right \rangle}}{\left \langle u,u \right \rangle} = \left \langle v,v \right \rangle$$

since inner-product is scalar hence we have

$\left \langle ku,v \right \rangle = \left \langle v,v \right \rangle$ for some scalar $k$

so now my question is, is this enough to conclude that $v = ku$ thus, proving the claim ?

else how should I approach this question ?

Any help or insight is deeply appreciated.

Answer 1

If $\vert\langle u,v\rangle\rvert=\Vert u\Vert\cdot\Vert v\Vert$, we first consider $v={\it 0}$, then the result holds trivially. Next, consider $v\neq{\it 0}$ and let $a=\frac{\langle u,v\rangle}{\Vert v\Vert^2}$, then \begin{align*} \Vert u-av\Vert^2 &=\Vert u\Vert^2-\bar{a}\langle u,v\rangle - a\langle v,u\rangle+a\bar{a}\Vert v\Vert^2\\ &=\Vert u\Vert^2-\frac{\overline{\langle u,v\rangle}}{\Vert v\Vert^2}\langle u,v\rangle - \frac{\langle u,v\rangle}{\Vert v\Vert^2}\overline{\langle u,v\rangle}+\frac{\langle u,v\rangle\overline{\langle u,v\rangle}}{\Vert v\Vert^2}\\ &=\Vert u\Vert^2-\frac{\vert\langle u,v\rangle\vert^2}{\Vert v\Vert^2} =\Vert u\Vert^2-\frac{\Vert u\Vert^2\cdot\Vert v\Vert^2}{\Vert v\Vert^2}=0. \end{align*} It follows that $u-av={\it 0}\,$ or $u=av$. Conversely, we consider $v={\it 0},$ then $$\vert\langle u,v\rangle\vert=\vert\langle u,{\it 0}\,\rangle\vert=0=\Vert u\Vert\cdot 0=\Vert u\Vert\cdot\Vert v\Vert.$$ which is trivial. Finally, consider $v\ne{\it 0}$, then we can write $u=cv$ for some scalar $c$. Thus $$\vert\langle u,v\rangle\vert=\vert\langle cv,v\,\rangle\vert=\vert c\vert \cdot\Vert v\Vert^2=(c\cdot\Vert v\Vert)\cdot\Vert v\Vert =\Vert cv\Vert\cdot\Vert v\Vert=\Vert u\Vert\cdot\Vert v\Vert.$$

Answer 2

Look at the proof of the Cauchy-Schwarz inequality:

$$0 \leq \langle u+\alpha v ,u+\alpha v \rangle = \Vert u\Vert^2+|\alpha|^2\Vert v\Vert^2+\alpha \langle v,u\rangle +\overline{\alpha}\langle u,v\rangle,$$

for $\alpha =\frac{-\langle u,v\rangle }{\Vert v\Vert^2}$. The only way we have equality, is if we have equality in the first "$\leq $". This gives $u+\alpha v=0$, or $u=-\alpha v$.

This answer is similar to the one just posted, but it highlights an important technique for problems like these: when one wants to determine when equality holds in inequalities, go back and look at the proof of the inequality and see what happens if the "$\leq$" signs were "$=$".