Prove that $\langle u,v\rangle=0\Longleftrightarrow \|u\|\leq \|u+av\|$.

So far I can get the $\Longrightarrow$ very easily, but I need some help with the $\Longleftarrow$ implication, any hints would be greatly appreciated. Please no answers, I just want a small nudge to get me in the right direction, I've been stumped on this for a little bit. Here's what I have so far:


Assume $\|u\| \leq \|u+av\|$

$\|u+av\|^2=\langle u+av,u+av\rangle=\langle u,u\rangle+ \langle u,av\rangle+\langle av,u\rangle+\langle v,v\rangle=\|u\|^2+\bar{a} \langle u,v\rangle+a\langle v,u\rangle+\|v\|^2=\|u\|^2+ \bar{a}\bar{\langle v,u\rangle}+a\langle v,u\rangle+\|v\|^2$


  • $\begingroup$ Let $u = (1, 0)$ and $v = (1,0)$ and $a = 1$. Then $\|u\| = 1; \|u + av\| = 2$, but $\langle u, v \rangle = 1$. That seems to contradict your conjecture. $\endgroup$ Feb 19 '14 at 23:47
  • 2
    $\begingroup$ Do you possibly mean that the left side holds for all $a$? $\endgroup$
    – robjohn
    Feb 19 '14 at 23:49
  • $\begingroup$ yes, for all a in the field. $\endgroup$ Feb 20 '14 at 1:26

Assuming that you mean that the inequality hold for all $a$, we have $$ \|u\|^2\le\|u+av\|^2\tag{1} $$ Let $a=t\langle u,v\rangle$ for $t\in\mathbb{R}$, then $(1)$ implies $$ \begin{align} \langle u,u\rangle &\le\langle u+av,u+av\rangle\\ &=\langle u,u\rangle+2\mathrm{Re}\left(a\,\overline{\langle u,v\rangle}\right)+|a|^2\langle v,v\rangle\\ &=\langle u,u\rangle+2t\left|\langle u,v\rangle\right|^2+t^2\left|\langle u,v\rangle\right|^2\langle v,v\rangle\\ 0&\le\left(2t+t^2\langle v,v\rangle\right)\left|\langle u,v\rangle\right|^2 \end{align} $$

Since the right hand side is $0$ when $t=0$, and the right side must always be greater than or equal to $0$, the derivative at $t=0$ must be $0$. That is, $$ \langle u,v\rangle=0 $$

  • $\begingroup$ @user104254: sorry, it was just a hint, but enough time has passed that it seems fine to adjust the answer for a complex vector space. $\endgroup$
    – robjohn
    Mar 10 '14 at 7:23
  • $\begingroup$ @user104254: It seemed that the issues around $\langle u,v\rangle=0$ seemed to complicate the argument. I guess it's worth considering. $\endgroup$
    – robjohn
    Mar 10 '14 at 11:54
  • $\begingroup$ It's potentially worth noting here that from the first equality, one can just compute the real part of $a \overline{\langle u,v\rangle}$ in terms of the real and imaginary parts of $a$ and $\langle u,v \rangle$. Then taking appropriate limits along the real and imaginary axes shows that the real and imaginary parts of $\langle u,v \rangle$ must both be $0$. This approach might be more intuitive for some than setting $a = t \langle u,v \rangle$ and using derivatives. $\endgroup$
    – Eli.Orvis
    Jan 30 at 17:54

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