Prove that if $x^TAx = 0 \Rightarrow Ax = 0$

Question

A is $n \times n$ symmetric and positive semidefinite. So we can use $A^T = A$ and $x^TAx \geq 0$ for all x $\in \mathbb{R}^n$.

Proof by contradiction comes to mind, or $x^TAx = 0 \; \land \; Ax \neq 0$. This might seem like a given at first but I'm thinking what if there is some linear dependence between $x^T$ and the $Ax$ vector?

Edit: Ideally, I'm supposed to prove this using the rank of a matrix.

If I were to prove it by contradiction, $x^TAx = 0 \; \land \; Ax \neq 0$, I'd start off by saying that:

• $x$ cannot be a zero vector in this case. The original statement $x^TAx = 0 \Rightarrow Ax = 0$ is trivial for a zero $x$ vector and needs no proof.
• from $Ax \neq 0$ we have $rank(Ax) = 1$ since $Ax$ is also a non-zero vector.
• problem: how to show that the $x^T$ vector and the $Ax$ vector cannot be orthogonal? (this would be the point of the contradiction proof)
• problem 2: the only idea I have for a proof barely utilizes the rank of a matrix, which is my assignment.

Answer 1

Consider the EVD of the matrix $A=UDU^T$.

Then we can rewrite $x^TAx=x^TUD^\frac12D^\frac12U^Tx=\|D^\frac12U^Tx\|^2$

Hence if $x^TAx=0$, we must have $D^\frac12Ux=0$ and hence $$UD^\frac12(D^\frac12U^Tx)=(UD^\frac12D^\frac12U^T)x=Ax=0$$

Answer 2

I think you have the best hint from Gerry Myerson in the link from @projectilemotion.

Symmetric $A \Rightarrow$ we can choose orthogonal eigenvectors $v$ for A. These form a basis for $\mathcal{R}^n$, so you can write $x=\sum b_jv_j$.

Positive semi-definite $A \Rightarrow$ its eigenvalues are non-negative. Express $x^TAx$ in terms of the eigenvector basis and use these two facts, you should be home.