Regarding spectral radius of a matrix

Question

Let $n\geq 2$, $s\leq n$ and $r_1,r_2,\cdots, r_s$ with $\sum_{j=1}^{s}r_j=n. $ Let $E$ be the subspace of ${\mathbb{C}}^{n\times n}$ given by the following scalar block diagonal matrices. $$E= \{diag[z_1I_{r_1},\cdots, z_sI_{r_s}]\in {\mathbb{C}}^{n\times n}: z_1,\cdots,z_s\in\mathbb{C}\}$$ Now for every $A\in {\mathbb{C}}^{n\times n}$ define the value $$\mu_E(A)=\frac{1}{\inf\{\|X\|:X\in E, \det(I_n -AX)=0\}}.$$ Let $B_{n\times n}=\{X\in {\mathbb{C}}^{n\times n}: \|X\|<1\}.$ Can anyone help me prove that for any $A\in {\mathbb{C}}^{n\times n}$, $\mu_E(A)=1$ if and only if $\max_{X\in \bar{B}_{n\times n}\cap E}\rho(XA)=1?$ Where $\rho$ stands for the spectral radius. I was able to prove that for a given $n$ if $s=1$ and $r_1=n$, then $\mu_E(A)=\rho(A)$.

Answer 1

Define $m=\inf\{\|Y\|:Y\in E,\,\det(I-AY)=0\}$ and $M=\max\left\{\rho(AX): X\in E,\,\|X\|\le1\right\}$.

Suppose $\{\|Y\|: Y\in E,\,\det(I-AY)=0\}$ is nonempty. Then $m>0$, because the determinant function is continuous and $\det(I)\ne0$. Also, $m$ is attainable, for, suppose $\|Y_j\|\to m$ for some sequence $\{Y_j\}_{j\in\mathbb N}$ of feasible $Y$s. Then $\{Y_j\}_{j\in\mathbb N}$ is bounded and it contains a subsequence that converges to some $Y_0$ in the closed space $E$. As the determinant function is continuous, we get $\det(I-AY_0)=0$. Hence $Y_0$ is a minimiser. Since $1$ is an eigenvalue of $AY_0$, we have $\rho(AY_0)\ge1$. Let $X'=\frac{Y_0}{\|Y_0\|}$. Then $X'\in E$ and $\|X'\|=1$. Therefore $$ M\ge\rho(AX')=\frac{\rho(AY_0)}{\|Y_0\|}\ge\frac{1}{\|Y_0\|}=\frac{1}{m}>0.\tag{1} $$

Conversely, suppose $M>0$. Then this maximum value is attained and attainable only on the unit sphere, because the unit ball in $E$ is compact and $X\mapsto\rho(AX)$ is positively homogeneous. Let $X_0=\arg\max\left\{\rho(AX): X\in E,\,\|X\|\le1\right\}$ with $\|X_0\|=1$. Define $Y'=\frac{e^{i\theta}X_0}{\rho(AX_0)}$. Then $Y'\in E$ and $1$ is an eigenvalue of $AY'$ for an appropriate choice of $\theta\in\mathbb R$. Therefore the set $\{Y\in E,\,\det(I-AY)=0\}$ is nonempty (because it contains $Y'$) and from our previous discussion, we know that $m>0$. Thus $$ 0<m\le\|Y'\|=\frac{\|X_0\|}{\rho(AX_0)}=\frac{1}{\rho(AX_0)}=\frac{1}{M}.\tag{2} $$

To sum up, $m>0$ if and only if $M>0$, and in case they are positive, the inequalities $(1)$ and $(2)$ show that $M=\frac{1}{m}$. The conclusion now follows.