Given $A \in \mathbb{R}^{n \times n}$ such that $y^T A y \geq 0$ for any $y \in \mathbb{R}^n$, does $x^TAx=0$ imply $Ax=0$?

Question

Given $A \in \mathbb{R}^{n \times n}$ such that $y^T A y \geq 0$ for any $y \in \mathbb{R}^n$, does $x^TAx=0$ imply $Ax=0$?

I think i have proven it in the case where $A$ is symmetric (so $A$ is positive semidefinite), but does it hold in this more general setting?

Answer 1

In other words, is it possible for a linear map to transform a nonzero vector $x$ into a vector that is orthogonal to $x$? Asked this way, the answer is trivially yes (that is, the answer to your initial question is no).

It's even possible for the linear map to transform every nonzero vector into an orthogonal one: for instance in two dimensions, a rotation of angle $\pi/2$.

$$A=\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}$$

Then

$$x^TAx=\begin{pmatrix}x_1 &x_2\end{pmatrix}\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix} \begin{pmatrix}x_1 \\ x_2\end{pmatrix}=\begin{pmatrix}x_1 &x_2\end{pmatrix}\begin{pmatrix}-x_2 \\ x_1\end{pmatrix}=0$$

Answer 2

it is only true for positive definite matricies, whose definition are exactly : $$x^TAx>0 \text{ if }x\neq 0 $$

This is explained by the fact that if the matrix is only positive semi definite, you only have : $$\lambda\in Sp(A)\Rightarrow \lambda \geq 0 $$