Proof of the theorem that there is no counterexample If $\rho(A) $ is a spectral radius and $ \|\cdot\|$ is norm, the relation $\rho(A) \leqslant \|A\| $ is always established.
I want to prove that if $\rho(A) <1$ then $ \|A\|<1$.
Note that using the computer, I checked a large number of random matrices and there were no Counterexample. This is probably a mathematical theorem, but I can not prove it. please help. Thankful
$$ $$
A friend found a counterexample for it so I change the condition to $0<\rho(A) <1$. Can the theorem be proved now?
 A: Suppose $ V^{-1} A V = J$ where $J$ is the Jordan normal form. Write
$J= \Lambda+ T$, where $\Lambda$ is diagonal and $T$ is strictly upper triangular.
Let $\|x\|_B = \|B^{-1} x\|$, where the latter is the Euclidean norm. It is easy to check that the corresponding induced norm is $\|A\|_B = \|B^{-1}A B \|$.
Note that if $A$ is diagonalisable (that is, $T =0$) then we see that
$\|A\|_V = \|\Lambda\| = \rho(A)$.
If $T \neq 0$ we need an adjustment. The idea is straightforward but the description is a little messy.
Let $\Sigma=\operatorname{diag}(\sigma_1,...,\sigma_n)$, suppose all $\sigma_k>0$.
Note that $\Sigma \Lambda \Sigma^{-1} = \Lambda$ and
$[ \Sigma T \Sigma^{-1}]_{ij} = {\sigma_i \over \sigma_j} [T]_{ij}$.
Note that the latter term is always zero for $i \le j$.
Pick some $\epsilon>0$ and let $\sigma_1 = 1$, we will choose the remaining
$\sigma_k$ such that $|{\sigma_i \over \sigma_j} [T]_{ij}| < \epsilon$.
Suppose we have chosen $\sigma_1,...,\sigma_i$, then pick some $t >0$
such that $|{\sigma_i \over t} [T]_{ij}| < \epsilon$ for $j=i+1,...,n$ and
let $\sigma_{j+1} = t$.
Then we have
$\| A\|_{V \Sigma^{-1}} = \|\Sigma V^{-1} AV\Sigma^{-1}\| = \|\Sigma (\Lambda+T)\Sigma^{-1}\| \le \rho(A)+\| \Sigma T\Sigma^{-1} \|$
and it is straightforward to check that $\| \Sigma T\Sigma^{-1} \| \le n \epsilon$.
