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July 29th, 2014, 10:21 AM   #1
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prove that $x$ and $y$ are multiples

Hellow!
I've got the following problem:
"Let $x,y \in \mathbb R^n$. If every $z \in \mathbb R^n$ which is orthogonal to $x$ is also orthogonal to $y$, prove that $x$ and $y$ are multiples".

Proof: consider two vectors $x,y$. The projection of $y$ into $x$ is $w=\frac{<x,y>x}{<x,x>}$. Then we have $<y-w,x>=0$. By hypothesis of the problem, $<y-w,y>=0 \Rightarrow |y|^2=<w,y>$. If I could get that $|x|^2|y|^2=<x,y>^2$, it's done. But how?
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July 29th, 2014, 01:15 PM   #2
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Let y = ax + z, where (x,z) = 0, with a = (x,y)/(x,x). However, since (x,z) = 0, then (y,z) = 0. Therefore (y,y) = [(x,y)]^2/(x,x), which is what you are trying to show.
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