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Let $A$ be a real $2\times2$ symmetric matrix that is indefinite (it has a positive and a negative eigenvalues). Let $Q = \langle Av, v\rangle$ be the associated quadratic form. If $Q(v) = 0$, is it true that $$|Av|^2 = -\det(A)|v|^2?$$

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Denote the eigenvalues of $A$ by $\lambda_1 < 0 <\lambda_2$ with corresponding eigenvectors $v_i$ with norm one. These are orthogonal: $v_1^Tv_2=0$, thus $\{v_1,v_2\}$ is an orthonormal basis. Then $v= v_1 (v^Tv_1) + v_2(v^Tv_2)$ for all $v$, and $Av=\lambda_1v_1 (v^Tv_1) + \lambda_2v_2(v^Tv_2)$. $Q(v)=0$ implies $0=v^TAv=\lambda_1 (v^Tv_1)^2 + \lambda_2(v^Tv_2)^2$. Then $$\begin{split} \|Av\|^2 &= \lambda_1^2 (v^Tv_1)^2 + \lambda_2^2 (v^Tv_2)^2\\ &= -\lambda_1\lambda_2(v^Tv_2)^2 - \lambda_2\lambda_1(v^Tv_1)^2\\ &= -\lambda_1\lambda_2 \|v\|^2 \\ &=-\det(A)\|v\|^2. \end{split}$$


The original question was without squaring the norms...

No, this is not true. If the claim would be true for $A$, it would be true for $2A$. But $\|2Av\|=2\|Av\|$ and $\det(2A)=4\det(A)$.

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  • $\begingroup$ You are indeed correct, I forgot to square the norm! $\endgroup$ – Henrique Augusto Souza Jul 8 at 11:31

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