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Let V be a vector space over $\mathbb{R}$. Let $(x,y) \mapsto \langle x,y\rangle$ be an inner product on $V$ with induced norm $\lVert x\rVert=\sqrt{\langle x,y\rangle}$. Suppose that $x$ and $y$ are two vectors in $V$ such that $\lVert x\rVert=\lVert y\rVert=1$ and $\langle x,y\rangle=1$. Show that $x=y$.

This looks really obvious, but I tried using $(\lVert x\rVert-\lVert y\rVert)^2$ and Cauchy inequality to approach it, and still didn't get it. I am running out of ideas now. Any help is appreciated.

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    $\begingroup$ Please use proper delimiters (\langle, \rangle, \lVert, \rVert) for the symbols. $\endgroup$ Jul 16 '20 at 19:57
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Notice that $$\|x-y\|^2 = \langle x-y,x-y \rangle = \langle x,x \rangle - 2 \langle x,y \rangle + \langle y,y \rangle = 1-2+1 = 0,$$ so $\|x-y\| = 0$ and hence $x=y$.

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Here's a bit more geometric an argument. Consider the projection of $x$ onto $y$. The usual formula gives $$\text{proj}_y x = \frac{\langle x,y\rangle}{\|y\|^2} y = \frac11 y = y.$$ Now the Pythagorean Theorem says that $x=y$. If we have $x=y+z$ with $z$ orthogonal to $y$, then $\|x\|^2 = \|y\|^2+\|z\|^2\ge \|y\|^2$; thus, $\|x\| = \|y\|$ if and only if $z=0$.

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  • $\begingroup$ Beautiful proof! $\endgroup$ Jul 16 '20 at 20:06
  • $\begingroup$ Thanks, @BrianBritosSimmari. I usually try to emphasize things geometric in linear algebra. But this exercise was a new one for me. $\endgroup$ Jul 16 '20 at 21:52
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You have

$$\Vert x-y \Vert^2 = \Vert x\Vert^2 -2\langle x,y \rangle + \Vert y \Vert^2=0$$

Hence $x=y$.

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By Cauchy-Schwarz inequality, \begin{align*} 1 = |\langle x, y \rangle| \leq \|x\|\|y\| = 1 \times 1 = 1. \end{align*} Hence the equality holds. Check the equality condition for Cauchy-Schwarz inequality: we know that the equality holds if and only if $y = \lambda x$ for some $\lambda$. Taking norms on both sides yields $\lambda = \pm 1$. But $\langle x, y \rangle = 1$ rules out the case $\lambda = -1$.

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