If $M\geq 0$, why $M$ is invertible if and only if $\left< Mv, v \right> = 0$ implies that $v = 0$? Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex finite dimensional Hilbert space $F$. Let $M\in \mathcal{B}(F)^+$ (i.e. $\langle Mx\;, \;x\rangle\geq 0$ for all $x\in F$.
I want to show that $M$ is invertible if and only if $\left< Mv, v \right> = 0$ implies that $v = 0$.
If $F$ is real finite dimensional Hilbert space. Is the above equivalence remains true?
Thank you!
 A: A positive-semidefinite operator $M$ is necessarily self-adjoint.
Since $F$ is finite-dimensional, there exists an orthonormal basis $\{e_1, \ldots, e_n\}$ for $F$ of eigenvectors for $M$. Assume the respective eigenvalues are $\lambda_1, \ldots, \lambda_n$. Since $M$ is self-adjoint, the eigenvalues are real, and since $M$ is positive-semidefinite, we have $\lambda_i \ge 0$.
For $v \in F$ we have:
\begin{align}
\langle Mv, v\rangle &= \left\langle M\left(\sum_{i=1}^n \langle v, e_i\rangle e_i\right), \sum_{i=1}^n \langle v, e_i\rangle e_i\right\rangle\\
&= \left\langle \sum_{i=1}^n \langle v, e_i\rangle Me_i, \sum_{i=1}^n \langle v, e_i\rangle e_i\right\rangle\\
&= \left\langle \sum_{i=1}^n \langle v, e_i\rangle \lambda_ie_i, \sum_{i=1}^n \langle v, e_i\rangle e_i\right\rangle\\
&= \sum_{i=1}^n \left|\langle v, e_i\rangle\right|^2\lambda_i
\end{align}
Assume $\langle Mx, x\rangle = 0 \implies x = 0$. To show that $M$ is invertible, it suffices to show $\operatorname{Ker}\, M = \{0\}$. Let $Mx = 0$. We have
$$\langle Mx, x\rangle = \langle 0, x\rangle = 0 \implies x = 0$$
Hence, $\operatorname{Ker}\, M = \{0\}$ so $M$ is invertible.
Now, assume that $M$ is invertible. This means that $\lambda_i > 0$ for all $i \in \{1, \ldots, n\}$.
For every $x \ne 0$ we have that $\left|\langle x, e_i\rangle\right| \ne 0$ for some $i$. Therefore:
$$\langle Mx, x \rangle = \sum_{i=1}^n \left|\langle v, e_i\rangle\right|^2\lambda_i \ge \left|\langle x, e_i\rangle\right|^2\lambda_i > 0$$
We conclude that $\langle Mx, x\rangle = 0 \implies x = 0$.
