# Lower bound for $x^TMx$ when $\lambda_{min}=0$

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?

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.
Denote the eigenvalues $\lambda_{min}=0$ then there exist nonzero eigenvector $v$ such $Mv=\lambda_{min}v$ and therefore $v^{\top}Mv=0 v^{\top}v = 0$.