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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?

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  • $\begingroup$ The statement cannot be proven because it is obviously incorrect for an arbitrary square matrix. Take for example matrix $W=\begin{bmatrix} 1 & 0 \\ 0 & -1\end{bmatrix}$ and vector $v = \begin{bmatrix} 1 \\ 1\end{bmatrix}$. Then $v^TWv=0$ and $Wv = \begin{bmatrix} 1 \\ -1\end{bmatrix}$ $\endgroup$ – Andre Zege May 5 '18 at 18:56
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    $\begingroup$ The question specifically states that W is positive semidefinite, @AndreZege, which your W is not. Your answer correctly address the thread title (which does not mention W being positive semidefinite), but not the body of the question. $\endgroup$ – Mark L. Stone May 5 '18 at 19:22
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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 $W v=0$.

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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

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  • $\begingroup$ Are you sure for any symmetric matrix $W$, there exists $W^{\frac{1}{2}}$ such that $W^{\frac{1}{2}}$ is also symmetric?? $\endgroup$ – Dony May 16 '17 at 3:51
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    $\begingroup$ @Dony Yes, your operator is positive (also referred to as positive semi-definite). $\endgroup$ – Matthew Gunn May 16 '17 at 7:45
  • $\begingroup$ The argument assumes that $W$ is positive definite, if only semi-definite not. $\endgroup$ – kjetil b halvorsen May 16 '17 at 8:06
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    $\begingroup$ @kjetilbhalvorsen What argument assumes positive definite? I've only assumed the operator is positive (which is not the same as positive definite). There's nothing wrong with eigenvalues of 0. The real spectral theorem (on a real inner product space) only requires that a linear operator is self-adjoint. $\endgroup$ – Matthew Gunn May 16 '17 at 8:14
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    $\begingroup$ @kjetilbhalvorsen Let $\mathbf{z} = W^{\frac{1}{2}} \mathbf{x}$. Observe $\mathbf{z}$ is a vector and I'm taking the usual $L_2$ norm $\|\mathbf{z}\|_2 = \sqrt{\sum_i z_i^2}$ of that vector. I'm not using $W$ to define a norm $\| . \|_W$ (and I agree with you, that would be a mistake because it would fail $\|\mathbf{x}\|_W=0$ iff $\mathbf{x} = \mathbf{0}$). $\endgroup$ – Matthew Gunn May 16 '17 at 8:54

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