Is the set of matrices with constrained condition numbers a convex set?

Let $\mathbb{S}_{\tau}^{+,p}$ indicates the $p\times p$ real symmetric positive-semidefinite matrix, whose condition number, defined as the ratio of maximum eigenvalue and minimum eigenvalue, is less or equal to $\tau$. Is $\mathbb{S}_{\tau}^{+,p}$ convex?

I attempted to use the definition of convexity to prove it.

Suppose $M_1,M_2\in \mathbb{S}_{\tau}^{+,p}$, and $\forall v\in(0,1)$, I would like to show $vM_1+(1-v)M_2\in\mathbb{S}_{\tau}^{+,p}$.

I understand that $$\lambda_{\text{min}}\left[\nu M_1+(1-\nu)M_2\right]\geq\lambda_{\text{min}}[\nu M_1]+\lambda_{\text{min}}[(1-\nu) M_2]\geq\min\{\lambda_{\text{min}}[M_1], \lambda_{\text{min}}[M_2]\}$$ and $$\lambda_{\text{max}}\left[\nu M_1+(1-\nu)M_2\right]\leq\lambda_{\text{max}}[\nu M_1]+\lambda_{\text{max}}[(1-\nu) M_2]\leq\max\{\lambda_{\text{max}}[M_1], \lambda_{\text{max}}[M_2]\},$$ but they do not guarantee that $$\frac{\lambda_{\text{max}}\left[\nu M_1+(1-\nu)M_2\right]}{\lambda_{\text{min}}\left[\nu M_1+(1-\nu)M_2\right]}\leq \tau$$

If the statement is not true, can you give a simple counterexample? Thanks!

Solution 1:

$\def\R{\mathbb{R}}\def\T{^{\mathrm{T}}}\DeclareMathOperator{\cond}{cond}\def\paren#1{\left(#1\right)}$Denote by $S_n$ the set of all $n × n$ symmetric positive-definite matrices over $\R$. Note that for any $A \in S_n$, its largest and smallest eigenvalues are given by $\paren{ \max\limits_{x\T x = 1} x\T Ax }^{\frac{1}{2}}$ and $\paren{ \min\limits_{x\T x = 1} x\T Ax }^{\frac{1}{2}}$, respectively. Thus$$ \cond(A) = \paren{ \frac{\max\limits_{x\T x = 1} x\T Ax}{\min\limits_{x\T x = 1} x\T Ax} }^{\frac{1}{2}},$$ and $$A \in S_n(τ) \iff \max_{x\T x = 1} x\T Ax \leqslant τ^2 \min_{x\T x = 1} x\T Ax.$$

Now for $A, B \in S_n(τ)$ and $t \in (0, 1)$, since\begin{align*} &\mathrel{\phantom=} \max_{x\T x = 1} x\T ((1 - t)A + tB)x = \max_{x\T x = 1} ((1 - t)x\T Ax + tx\T Bx)\\ &\leqslant \max_{x\T x = 1} (1 - t)x\T Ax + \max_{x\T x = 1} tx\T Bx = (1 - t)\max_{x\T x = 1} x\T Ax + t\max_{x\T x = 1} x\T Bx\\ &\leqslant (1 - t)τ^2 \min_{x\T x = 1} x\T Ax + tτ^2 \min_{x\T x = 1} x\T Bx = τ^2 \paren{ \min_{x\T x = 1} (1 - t)x\T Ax + \min_{x\T x = 1} tx\T Bx }\\ &\leqslant τ^2 \min_{x\T x = 1} ((1 - t)x\T Ax + tx\T Bx) = τ^2 \min_{x\T x = 1} x\T ((1 - t)A + tB)x, \end{align*} then $(1 - t)A + tB \in S_n(τ)$.

Remark: As explained in the comments above, $S_n(τ)$ contains only positive-definite matrices since for any positive-semidefinite $A$, if $0$ is an eigenvalue of $A$, then the condition number of $A$ is ill-defined.