Show the the equivalence between $v^H W v = 0$ and $W v=0$ Can someone help me to prove the equivalence between $v^{H}W v = 0$ and $W v=0$, where $W$ is positive-semidefinite?
I can finish the part: $W v=0 \rightarrow v^HWv = 0$.
But the inverse part: $v^HWv = 0 \rightarrow Wv=0$, I cannot find ways to prove. 
Can anyone help me to show the inverse part? 
 A: Assuming $W$ is positive semidefinite, it has orthogonal eigenvectors $w_{i}$ and corresponding non-negative eigenvalues $\lambda_{i}$. Then, from expanding $v$ into the basis of eigenvectors $v=\sum \alpha_{i} w_{i}$ one finds $Q=v^T W v = \sum \alpha_{i}^{2} \lambda_{i}$. From $Q=0$, follows that $\alpha_{i}=0$ for all $\lambda_{i} > 0$. Only $\alpha_{i}$ corresponding to $\lambda_{i}=0$. could be non-zero. Hence $W v=0$.
A: The statement $ W \mathbf{v} = \mathbf{0}$ says that vector $\mathbf{v}$ belongs to the null space of $W$.
Let's assume for a moment a symmetric transformation $W^\frac{1}{2}$ exists. Then the statement $\mathbf{v}' W \mathbf{v} = 0$ is equivalent to saying
$ \|W^\frac{1}{2} \mathbf{v} \| = 0$, which (by def. of a norm)  is the same as as saying $W^\frac{1}{2} \mathbf{v} = \mathbf{0}$ and $\mathbf{v}$ belongs to the null space of $W^\frac{1}{2}$.
The existence of a self-adjoint $W^\frac{1}{2}$ with the same null space as $W$ can be shown via the spectral theorem.
The spectral theorem allows you to write $W$ as:
$$ W = P D P'$$
Where $D = \begin{bmatrix} \lambda_1 & 0 & \ldots \\ 0 & \lambda_2 & \ldots \\ \ldots \end{bmatrix}$ is a diagonal matrix of eigenvalues and $P$ is an orthogonal matrix (whose columns are eigenvectors). Then $W^\frac{1}{2} = P D ^ \frac{1}{2} P'$. The eigenvectors with zero eigenvalues are the same for $W$ and $W^\frac{1}{2}$ hence they have the same null space.
References
Axler, Sheldon. Linear Algebra Done Right, 1991
