# Determinants and indefinite symmetric matrices

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 $$|Av|^2 = -\det(A)|v|^2?$$

## 1 Answer

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 $$\det(2A)=4\det(A)$$.

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