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For a set of linear independent vectors $x_1,x_2,...,x_n$, we know that if$$ k_1 * x_1+k_2 * x_2 + ... + k_n * x_n = 0$$ Then, $k_1=k_2=...=k_n = 0$.

Now I want to extend this. Suppose that vectors $x_1,x_2,...,x_n$ are linear independent. The constraint $$ \sum_{i=1}^{n} x_{i}^{T}Ax_{i} = 0$$ with a positive semidefinite $A$ will lead to a result that $Ax_i = \mathbf{0},\forall i$.

This can help prove that if $Trace(AX) = 0$, so that $Ax_i = \mathbf{0}, \forall i$, where A,X are both positive semidefinite and $x_i$ is the eigenvector corresponding to the $i$th nonzero eigenvalue of matrix $X$. Can anyone help me prove it? Thank you.

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It's not true. Suppose my $x_1=\begin{pmatrix}1\\0\end{pmatrix}$ and $x_2=\begin{pmatrix}0\\1\end{pmatrix}$. I can choose $A=\begin{pmatrix}0&&1\\1&&0\end{pmatrix}$ I have $A x_1=x_2$, $A x_2=x_1$ so $x_i^T A x_i=0$ but $Ax_i\ne0$

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  • $\begingroup$ Thank you, you are correct, it is my negligence. There is also an additional condition that A is also positive semidefinite. So with this condition, can you prove the statement? $\endgroup$ – Ben Wu Apr 26 '16 at 22:45
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Guys I think I got the answer. Just for anyone who may need it.

Given that $A$ is positive semidefinite and $\sum_{i=1}^{N}x_{i}^{T}Ax_{i} = 0$, from definition we know that $x_{i}^{T}Ax_{i} = 0, \forall i$. Suppose that $A$ has $k$ nonzero eigenvalues. Then we do eigenvalue decomposition to $A$:

$$x_{i}^{T}Ax_{i} = x_{i}^{T}\sum_{j=1}^{k}a_{j}\lambda_{j}a_{j}^{T}x_{i}=\sum_{j=1}^{k}\lambda_{j}(a_{j}^{T}x_{i})^{2} = 0 \rightarrow a_{j}^{T}x_{i} = 0, \forall j \rightarrow a_{j}\lambda_{j}a_{j}^{T}x_{i} = \mathbf{0} ,\forall j\\\rightarrow \sum_{j=1}^{k}a_{j}\lambda_{j}a_{j}^{T}x_{i} = \mathbf{0} \rightarrow Ax_i = \mathbf{0}$$

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