bounded spectral radius $\Longrightarrow$ bounded matrix norm? There's a result in Linear Algebra that says something to the effect that

If a square matrix $A$ is such that $\rho(A)<1$, then there is some matrix norm such that $|||A|||<1$.

As an example, consider the matrix
$$
A:=\left(\begin{array}{cc}\frac{1}{2} & \frac{1}{16}\\ 1 & \frac{1}{2}\end{array}\right),
$$
with $\rho(A)=\frac{3}{4}<1$. The maximum column sum matrix norm of $A$ is given by
$$
{|||A|||}_1=\max\limits_{1\leq j\leq 2}\left(\sum\limits_{i=1}^{2}|a_{ij}|\right)=\frac{3}{2}>1,
$$
and the maximum row sum matrix norm is
$$
{|||A|||}_{\infty}=\max\limits_{1\leq i\leq 2}\left(\sum\limits_{j=1}^{2}|a_{ij}|\right)=\frac{3}{2}>1.
$$
My question is: which matrix norm is such that $|||A|||<1$, in view of the fact that $\rho(A)<1$? How does one go about figuring out such a norm, systematically?
 A: Suppose $A$ is a complex matrix with $\rho(A)<1$. Let $J=P^{-1}AP$ be the Jordan form of $A$. Let $D=\operatorname{diag}(1,\epsilon,\epsilon^2,\ldots,\epsilon^{n-1})$. Then the super-diagonal entries of $D^{-1}JD$ can be made arbitrarily small when $\epsilon>0$ is sufficiently small. It follows that $\lim_{\epsilon\to0}D^{-1}P^{-1}APD$ is the diagonal part of $J$ and hence $\|D^{-1}JD\|_\infty<1$ when $\epsilon$ is small. Now define a norm by $\|X\|=\|D^{-1}P^{-1}XPD\|_\infty$. Then $\|A\|<1$.
You may also define the norm as $\|D^{-1}P^{-1}XPD\|_1$ or $\|D^{-1}P^{-1}XPD\|_2$ in the above, provided that $\epsilon$ is sufficiently small. In your case, since $J=P^{-1}AP=\operatorname{diag}(\frac14,\frac34)$ for $P=\pmatrix{-\frac14&\frac14\\ 1&1}$, you may simply take $D=I$ and define $\|X\|=\|P^{-1}XP\|_p$ where $p=1,2$ or $\infty$.
A: There is an abstract way (similar to what is known as the 'Mather trick' in dynamical systems) to obtain an operator norm with the desired properties, using simply the definition of spectral radius and without the need for diagonalizing.
Let $(E,|\cdot|)$ be a Banach space (which may be infinite dimensional)  and $A\in L(E)$. If $\rho(A)<r<1$ then by the definition of spectral radius there is $C=C_r<+\infty$ so that:
$$ |A^n x| \leq C r^n |x|, \ \ n\geq 0, x\in E .$$
Now, let $r< \theta <1$ and define the 'adapted' norm:
$$ \|x \|_A = \sum_{n\geq 0} \frac{1}{\theta^n} |A^n x|. $$
Then $|x|\leq \|x\|_A \leq \frac{C}{1-r/\theta} |x|$ (so the norms are equivalent) and:
$$ \|A x\|_A = \sum_{n\geq 0} \frac{1}{\theta^n} |A^{n+1} x|\leq \theta \|x\|_A.$$
