Show x+y & x-y are Orthogonal if x & y Have Same Norms

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Show that x+y and x-y are orthogonal if and only if x and y have the same norms.

Can someone get me started?
 
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1) What does "orthogonal" mean here?

2) So if x+y and x-y are orthogonal what must be true?

3) And in order for that to be true what about x and y?
 
Can you think of a nice geometrical application of this result

\left\langle x+y, x-y\right\rangle =0 \Longleftrightarrow ||x||=||y|| ?

Daniel.
 
So (x+y)(x-y)=0, which can be turned into ||x||^2 = ||y||^2 take the square root of each side, I get ||x|| = ||y||.

As for dextercioby's question, if that is true, then x, y, and x+y make up an isosceles right triangle?
 
The paralelelogramme with perpendicular (onto another) diagonals is a rhombus. Therefore, the vectors have equal modulus. Actually, u've proven the reverse, viz.the geometrical result (theorem/proposition) by algebraic methods only. :wink:

Daniel.
 

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