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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?

Jimmy R
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2 Answers2

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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...

DonAntonio
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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
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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 $$\rm{Re}\langle \mathbf{x},\ \mathbf{y}\rangle \le|\langle \mathbf{x},\ \mathbf{y}\rangle|$$

EuYu
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