Is $x^TAx = y^TAy$ if $\|x\| =\|y\|$

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!

• 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 $A$? – amd Oct 13 '16 at 8:43
• No, that's all there is but than I must have gotten something else wrong along the way. Thank you guys! – Jonasson Oct 13 '16 at 10:26

No it is not correct, just consider $A=\left[\begin{matrix} 2 & 0 \\ 0 & 1\end{matrix} \right]$ and the two canonical basis vectors.

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 $v,w$.