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## Homework Statement

Let ##x,y \in \mathbb{R^n}## not null vectors. If for all ##z \in \mathbb{R^n}## that is orthogonal to ##x## we have that ##z## is also orthogonal to ##y##, prove that ##x## and ##y## are multiple of each other.

## Homework Equations

We can use that fact that ##<x , y-\frac{<x,y>}{|x|^2}x> = 0##

## The Attempt at a Solution

If ##x=y## the result follow directly. So let us suppose that ##x\neq y##.

Using the fact that ##<x , y-\frac{<x,y>}{|x|^2}x> = 0## and the hypothesis we can affirm that ##<y , y-\frac{<x,y>}{|x|^2}x> = 0##.

Further, by the fact that ##<x , y-\frac{<x,y>}{|x|^2}x> = 0## and ##<y , y-\frac{<x,y>}{|x|^2}x> = 0## we have:

$$<x , y-\frac{<x,y>}{|x|^2}x> = <y , y-\frac{<x,y>}{|x|^2}x> $$

$$\Rightarrow <x , y-\frac{<x,y>}{|x|^2}x> - <y , y-\frac{<x,y>}{|x|^2}x> = 0$$

$$\Rightarrow <x - y , y-\frac{<x,y>}{|x|^2}x> =0 \quad (*)$$

Until here everything seems great. But now...

Once ##x\neq y## , second the source I got this solution, by ##(*)## we can affirm that:

$$y-\frac{<x,y>}{|x|^2}x = 0$$

$$\Rightarrow y = \alpha x$$

Where ##\alpha = \frac{<x,y>}{|x|^2}x ##

But the fact is if we have ##<x,y>=0## not necessarily we must have ##x=0## or ##y=0##.

Can anyone see if I could not see a relevant fact in the argument or someone have another look for this problem?

Thanks in advance.

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