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Cauchy-Schwarz on real vector space: $| \langle u, v \rangle | = \| u \| \cdot \| v \| \iff u =\frac {\langle u, v \rangle} {\langle v, v \rangle} v$

Proving that $u =\frac {\langle u, v \rangle} {\langle v, v \rangle} v$ imply equality is easy by substitution of $u$.

However, I cannot prove that equality holds imply $u =\frac {\langle u, v \rangle} {\langle v, v \rangle} v$.

Can someone help me out ?

Fly by Night
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Shuzheng
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1 Answers1

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We have that $$ \lvert\langle u,v\rangle\rvert=\|u\|\|v\|, $$ iff $u$ and $v$ are linearly dependent.

Proof. In both are equal to 0, then they are linearly dependent. Assume that $v\ne 0$ and set $$ f(\lambda)=\|\lambda v+u\|^2. $$ Clearly $f(\lambda)\ge 0$, for all $\lambda\in\mathbb R$. But $$ \|\lambda v+u\|^2=\langle \lambda v+u,\lambda v+u\rangle=\|v\|^2\lambda^2+2\langle v,u\rangle \lambda+\|u\|^2, $$ which means that the roots of the quadratic polynomial above are either equal or complex or equivalently its discriminant is non-positive: $$ 0\ge \Delta=4\langle v,u\rangle^2-4\|u\|^2\|v\|^2. $$ Equality above implies that $$ \|\lambda v+u\|^2, $$ for some $\lambda\in\mathbb R$, which means that $u=-\lambda v$. Hence...

Yiorgos S. Smyrlis
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