Let $A$ be some matrix, If $x^TAx = 0$ for every vector $x$, then $A$ is singular. How can I prove it?
Here is what I thought:
first of all, for every matrice $M$, $$ x^T(M + M^T)x = x^TMx + x^TM^Tx = $$$$x^TMx + (x^TM^Tx)^T = x^TMx + x^TMx = 2(x^TMx) $$ $\rightarrow \frac{1}{2}x^T(M + M^T)x = (x^TMx)$
now since $x^TMx = 0$ for every x , $ \frac{1}{2}x^T(M + M^T)x = 0$ for every $x$. thus, $M + M^T = 0 \rightarrow -M = M^T$
Meaning $M$ is a skew-symmetric matrix, thus it's singular for sure.
Is it correct?