# Convexity of set $\lambda_{\min}(M) \geq a$ in the space of symmetric matrices

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.

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 $$q:=1-p$$

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

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

is a spectrahedron and, thus, convex.