Let $M$ be a symmetric matrix and $a \in \mathbb{R}$, $a \geq 0$. How do I go about showing that the set $$C = \left\{ \lambda_{\min}(M) \geq a : M \, \text{symmetric} \right\}$$ is a convex set?

If we restrict ourselves to positive definite matrices, I had the following idea:

$$\lambda_{\min}(M) \geq a \iff \lambda_{\min}(M)^{-1} \leq a^{-1} \iff \lambda_{\max}(M^{-1}) \leq a^{-1}$$

Now, since $\lambda_{\max}$ is a norm of symmetric matrices, the above set is a ball, hence convex. But I can't prove it for general symmetric matrices, which aren't necessarily invertible. Any help is greatly appreciated.


2 Answers 2


This comes down to the fact that two choices are better than one. First recall: $\min_{\mathbf x_: \Vert \mathbf x\Vert_2=1}:\text{trace}\big(\mathbf x\mathbf x^T M\big)=\min_{\mathbf x_: \Vert \mathbf x\Vert_2=1}:\mathbf x^T M\mathbf x = \lambda_\min(M)$

Thus for arbitrary $M, M' \in C$ and $p \in [0,1]$ and $q:=1-p$

$\lambda_\min\big(pM +qM'\big)$
$= \min_{\mathbf x_: \Vert \mathbf x\Vert_2=1}:\text{trace}\Big(\mathbf x\mathbf x^T \big(pM +qM'\big)\Big)$
$=\min_{\mathbf x_: \Vert \mathbf x\Vert_2=1}:\Big\{\text{trace}\Big(\mathbf x\mathbf x^T \big(pM\big)\Big)+\text{trace}\Big(\mathbf x\mathbf x^T \big(qM'\big)\Big)\Big\}$
$=\min_{\mathbf x, \mathbf y: \Vert \mathbf x\Vert_2=1 \& \Vert \mathbf y\Vert_2=1 \&\mathbf x:=\mathbf y}: \Big\{\text{trace}\Big(\mathbf x\mathbf x^T \big(pM\big)\Big)+\text{trace}\Big(\mathbf y\mathbf y^T \big(qM'\big)\Big)\Big\}$
$\geq \min_{\mathbf x, \mathbf y: \Vert \mathbf x\Vert_2=1 \& \Vert \mathbf y\Vert_2=1 }: \Big\{\text{trace}\Big(\mathbf x\mathbf x^T \big(pM\big)\Big)+\text{trace}\Big(\mathbf y\mathbf y^T \big(qM'\big)\Big)\Big\}$
$= p\cdot \min_{\mathbf x:\Vert \mathbf x\Vert_2=1 }: \text{trace}\Big(\mathbf x\mathbf x^T M\Big)+q\cdot \min_{\mathbf y:\Vert \mathbf y\Vert_2=1 }:\text{trace}\Big(\mathbf y\mathbf y^T \big(M'\big)\Big)$
$\geq p\cdot a +q\cdot a$
$\implies \big(pM +qM'\big) \in C$


$$ \mathcal S : = \left\{ {\bf X} \in \Bbb S_n (\Bbb R) \mid \lambda_{\min} ({\bf X}) \geq \alpha \right\} = \left\{ {\bf X} \in \Bbb S_n (\Bbb R) \mid {\bf X} \succeq \alpha \, {\bf I}_n \right\}$$

is a spectrahedron and, thus, convex.


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