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I have a symmetric matrix A and the vectors x,y. I'm struggling to show that $x^TAx = y^TAy$ if $\|x\|=\|y\|$ for example taking the 2-norm, could anyone help me out here?

Thanks a lot!

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  • $\begingroup$ You might be having difficulty because as stated, this is untrue. Take, for instance, $A=diag(2,1)$, $x=(1,0)$ and $y=(0,1)$. Are there any other conditions on $A$? $\endgroup$ – amd Oct 13 '16 at 8:43
  • $\begingroup$ No, that's all there is but than I must have gotten something else wrong along the way. Thank you guys! $\endgroup$ – Jonasson Oct 13 '16 at 10:26
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No it is not correct, just consider $A=\left[\begin{matrix} 2 & 0 \\ 0 & 1\end{matrix} \right]$ and the two canonical basis vectors.

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Take the case of $2\times2$ symmetric matrix that is singular. Then the equation $v^TAv=0$ can be translated ( $x$ and $y$ as real variables, not vectors), $ax^2+bxy+cy^2=0$. This is a conic. For example $x^2-2xy+y^2=0$, will represent the double line). It has infinitely many points/solutions.

So $v^TAv=w^TAw (=0)$ happens for distinct vectors $v,w$.

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