# Triangle equality implies vector dependence.

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?

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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. –  Jimmy R Oct 28 '12 at 19:24

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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I was thinking along the same lines but what is the concept of $\theta$ in N dimensions? –  Inquest Oct 28 '12 at 19:16
"Exactly" (sort of) the same as in two: take two vectors and the common plane containing then and there you know what this means. –  DonAntonio Oct 28 '12 at 19:20
Yes, but I only have the real part of $\langle x,y\rangle$ in the equality. –  Jimmy R Oct 28 '12 at 19:23
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 $$\operatorname{Re}\,\langle x,y\rangle=\frac{\langle x,y\rangle}{\cos\theta}\Longrightarrow \cos\theta =1\Longrightarrow \theta=0\Longrightarrow\,\,\,Q.E.D.$$ –  DonAntonio Oct 28 '12 at 19:31