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If $M$ is a positive semi-definite matrix, we know that $x^TMx \geq \lambda_{min} x^T x$. But if $\lambda_{min}=0$, is there any lower bound in terms of the smallest non-zero eigenvalue?

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If $\lambda_{min}=0$, then by definition $Mx=0$ for the corresponding eigenvector $x$. So $x^T M x=0$ and no better lower bound is possible.

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  • $\begingroup$ Of course! Thank you (and to iiivooo too). $\endgroup$ – Anush Mar 18 '16 at 16:11
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Denote the eigenvalues $\lambda_{min}=0$ then there exist nonzero eigenvector $v$ such $Mv=\lambda_{min}v$ and therefore $v^{\top}Mv=0 v^{\top}v = 0$.

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