Hermitian positive definite matrix defines an inner product on a vector space V

Question

my question is this:

I have to prove that for a positive-definite hermitian matrix $M$, i.e. such that $M=(m_{i,j})=(\overline{m_{j,i}})=\overline{M}^T$ of size $n$ x $n$, over the complex numbers, the following operation $$\langle x, y\rangle=\bf x^TM\bar{\bf y}$$ defines an inner product on the vector space. I have shown conjugate symmetry, and linearity in the first argument, and the fact that $\langle x, x\rangle\gt0$ for $x \neq 0$, as well as $x=0 \implies \langle x,x\rangle=0 $. I just need to show that if $$\langle x, x \rangle=0 \implies \bf {x}=0$$

I tried to reason it out simply by writing out $0=\langle x,x\rangle$ and expanding this according to the rule given above, but I don't see why it is that all the entries in the vector $\bf x$ have to be zero. I do know that the matrix must be nonzero, but beyond this I don't really know what to do.

Thanks for your help!

Answer 1

If $M$ is a positive-definite Hermitian matrix then

$$ x^* M x > 0 $$

for all non-zero complex vectors $x$ and $x^*$ means the conjugate-transpose of $x$. That means that $\langle x , x \rangle = 0$, iff $x = 0$

Answer 2

Express $M=NN^*,$

$$\langle x, x \rangle = x^TNN^*\bar{x}=\|N^*x\|^2=0$$

Hence $N^*x=0$ which implies that $x=0$.