Spectral radius of a matrix in a Banach algebra 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\}$.

*

*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 $A$.


*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!
 A: 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).

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 $.
