Show $A^T U A=0$ iff $A=0$ when $U$ is positive semidefinite matrix

Question

Let $U$ a symmetric,positive definite matrix. It's known that $A^T UA$ is a positive semidefinite matrix. My goal is to show that

$A^T U A =0$ iff $A=0$

The $\Longleftarrow$ implication is trivial.

The $\Longrightarrow$ is tricky to me: I'll try to show it by contradiction: suppose $A \ne 0$. Then, given a non-zero $v$ we have $Av = w \ne 0$

This implies that $v^TA^TUAv =w^TUw >0$ since U is S.P.D. How can I conclude that the whole matrix $A^T U A\ne 0$ ?

Any help is highly appreciated.

Answer 1

You don't need to consider the contrapositive. If $A^TUA=0$, then $x^TA^TUAx$ is identically zero. Since $U$ is positive definite, we obtain $Ax=0$. But $x$ is arbitrary. Hence $A$ must be zero.