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I am trying to prove this statement:

Show that if $x$ and $y$ are two vectors in an inner product space such that $||x+y||=||x||+||y||$, then $x$ and $y$ are linearly dependent.

Squaring the equality I get

$\langle x+y,x+y\rangle=\langle x,x\rangle +2||x||\cdot||y||+\langle y,y\rangle $ then, using linearity of the inner product I get

$ \langle x,x\rangle +\langle y,y\rangle+\langle x,y\rangle+\langle y,x\rangle=\langle x,x\rangle +2||x||\cdot||y||+\langle y,y\rangle $

After all the cancellation I finally arrive at

$ \mathrm{Re}\langle x,y\rangle=||x||\cdot||y|| $

This looks like Cauchy-Schwarz inequality, so the only thing left to show is that $\mathrm{Re}\langle x,y\rangle=|\langle x,y\rangle|$, how can I do that?

2 Answers 2

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Note that through an application of Cauchy-Schwarz we get $\rm{Re}\langle \mathbf{x},\ \mathbf{y}\rangle = \|\mathbf{x}\|\|\mathbf{y}\|\ge|\langle \mathbf{x},\ \mathbf{y}\rangle|$ This is only possible if there is equality since we naturally have $\rm{Re}\langle \mathbf{x},\ \mathbf{y}\rangle \le|\langle \mathbf{x},\ \mathbf{y}\rangle|$

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    Thanks, that's what I was looking for.2012-10-28
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But

$\langle x,y,\rangle=\left(||x||\cdot||y||\right)\cos\theta$

where $\,\theta=\,$the angle between the vectors $\,x,y\,$ , so...

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    Perhaps I'm missing something but I don't see the problem: you already had $\operatorname{Re}\,\langle x,y\rangle=||x||\cdot ||y||$ But according to the above we then get $\operatorname{Re}\,\langle x,y\rangle=\frac{\langle x,y\rangle}{\cos\theta}\Longrightarrow \cos\theta =1\Longrightarrow \theta=0\Longrightarrow\,\,\,Q.E.D.$2012-10-28