Spectral radius of a matrix in a Banach algebra

Question

Let $A$ be a $n \times n $ matrix.

Let $S(A)= \{ T: T= S^{-1} A S\ , $ for some invertible $ n\times n $ matrix $ S\}$.

1. Show that there exists $T \in \overline{S(A)}$ such that $\| T\|= \rho(T)= \rho(A) $, where $\rho(A)= \sup \{ |\lambda| : \lambda \in \sigma(A)\}$ is the spectral radius of $A$.

2. Is it possible to find $T \in S(A)$ such that $\| T\|= \rho(T)= \rho(A) $?

Things I know:

We know that the set of $n \times n$ matrices, $\mathbb{M}_{n \times n}$, can be identified as a Banach algebra.

Let $A \in \mathbb{M}_{n \times n}$.

By the spectral radius formula, $\rho(A) = \lim_{n \rightarrow \infty}\| A ^n\|^{1/n}$.

I'm not sure how to construct such $T \in \overline{S(A)}$, as I'm not sure what the closure of $S(A)$ is like.

Thank you in advance!

Answer 1

Without loss of generality we may assume that $A$ is in canonical Jordan form, namely $$ A = \pmatrix{ A_1 & 0 & \ldots & 0 \cr 0 & A_2 & \ldots & 0 \cr \vdots & \vdots & \ddots & \vdots \cr 0 & 0 & \ldots & A_k }, $$ where each $A_i$ is a Jordan block, that is $$ A_i=\pmatrix{ \lambda _i & 0 & 0 & \ldots & 0 & 0 \cr 1 & \lambda _i & 0 & \ldots & 0 & 0 \cr 0 & 1 & \lambda _i & \ldots & 0 & 0 \cr \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \cr 0 & 0 & 0 & \ldots & \lambda _i & 0 \cr 0 & 0 & 0 & \ldots & 1 & \lambda _i } $$ The Jordan decomposition is not guaranteed to take place over an orthonormal basis but, by conjugating $A$ with a matrix sending its Jordan basis to the canonical basis of $\mathbb C^n$, we may assume that $A$ is indeed in Jordan form relative to the orthonormal canonical basis.

Choose a parameter $\alpha >0$ and, for each Jordan block $A_i$, consider the invertible matrix $$ U^\alpha _i= \pmatrix{ \alpha & 0 & \ldots & 0 \cr 0 & \alpha ^2 & \ldots & 0 \cr \vdots & \vdots & \ddots & \vdots \cr 0 & 0 & \ldots & \alpha ^p }, $$ where $p$ is the size of $A_i$, and define $U^\alpha $ to be the block-diagonal matrix $$ U^\alpha = \pmatrix{ U^\alpha _1 & 0 & \ldots & 0 \cr 0 & U^\alpha _2 & \ldots & 0 \cr \vdots & \vdots & \ddots & \vdots \cr 0 & 0 & \ldots & U^\alpha _k }. $$ We then have that $U^\alpha A(U^\alpha )^{-1}$ is the block diagonal matrix with blocks $$ U^\alpha _iA_i(U^\alpha _i)^{-1} = \pmatrix{ \lambda _i & 0 & 0 & \ldots & 0 & 0 \cr \alpha & \lambda _i & 0 & \ldots & 0 & 0 \cr 0 & \alpha & \lambda _i & \ldots & 0 & 0 \cr \vdots & \vdots & \ddots & \ddots & \vdots & \vdots \cr 0 & 0 & 0 & \ldots & \lambda _i & 0 \cr 0 & 0 & 0 & \ldots & \alpha & \lambda _i }, $$ where one should note that the 1's in $A_i$ have been changed to $\alpha $'s.

Taking the limit of $U^\alpha A(U^\alpha )^{-1}$ as $\alpha \to 0$, we have that the limit matrix lies in the closure of $S(A)$ and it has all of the desired properties, including the fact that its spectrum is the same as that of $A$ (with different multiplicities).

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The answer to question 2 is no, a counter example being $$ A = \pmatrix{ 1 & 0 \cr 1 & 1 }. $$ The reason is that $$ N:= A-I $$ is a nonzero nilpotent matrix, and we claim that for any invertible matrix $S$, one has that the matrix $T:= SAS^{-1}$ has norm strictly bigger than 1. To see this, notice that $$ T = SAS^{-1} = S(I+N)S^{-1} = I+SNS^{-1} = I+S', $$ where $S'=SNS^{-1}$ is likewise a nonzero nilpotent matrix.

Arguing by contradiction, if $\|T\|\leq 1$, then also $\|T^k\|\leq 1$, for all $k$, but $$ T^k = (I+S')^k = I+kS', $$ whose norm tends to infinite as $k\to \infty $.