Proving result on spectral radius How do I prove that $$\rho(A)=\inf\limits_{\text{operator norms}}\|A\|,$$ $\rho$ being the spectral radius, $A$ being a complex $n\times n$ matrix and operator norms being induced from vector norms by defining $\|A\|=\min\limits_{\|x\|=1}\frac{\|Ax\|}{\|x\|}$? I have proved that $\rho(A)\leq\|A\|$ for any induced norm $\|\cdot\|$, so $\leq$ holds in the above equality. How do I prove $\geq$? 
 A: Let's be precise about what "proving $\geq$" entails.  What you need to show is that for any $\epsilon > 0$, there exists a matrix norm $\|\cdot\|$ such that
$$
\|A\| < \rho(A) + \epsilon
$$
The trick then, is to tailor a suitable example $\|\cdot \|$ to your matrix $A$ and choice of $\epsilon$.
I suggest the following: given the vector norm $|\cdot|$ of your choice, we can define the new vector norm
$$
\|x\| = |Sx|
$$
for any choice of invertible matrix $S$.  Consider the resulting induced norm, under a clever choice of $S$.

Further hint:
Note that the induced norm can be written as $\|A\| = |SAS^{-1}|$, since
$$
\max_{\|x\| = 1} \|Ax\| = 
\max_{|Sx| = 1}|SAx| =
\max_{|y| = 1}|SAS^{-1}y| = |SAS^{-1}|
$$

Further hint:
Note the following similarity:
$$
\pmatrix{
\lambda&1\\
&\lambda&\ddots\\
&&\ddots&1\\
&&&\lambda} \sim
\pmatrix{
\lambda&1/n\\
&\lambda&\ddots\\
&&\ddots&1/n\\
&&&\lambda}
$$
Consider using the $\|\cdot \|_1$ norm; any should work here, though.
A: I came here looking for an answer to this question for compact operators on Banach spaces. Here is a contruction that works in finite or infinite dimensions, assuming the spectral radius formula: $\rho = \lim_{n\rightarrow \infty} \|A^n\|^{1/n}  $. 
Define a norm in the underlying space as $ N(x) = \sum_{n \ge 0} \frac{\|A^nx\|}{(\rho +\epsilon)^n} .$ Trivially, $\|x\| \le N(x).$
From the spectral radius formula, given $\epsilon > 0$, there is a constant $C>0$ such that $\|A^n\| \le C(\rho + \epsilon/2)^n $ for all $n$. Then, 
$$ \sum_{n \ge 0} \frac{\|A^nx\|}{(\rho +\epsilon)^n} \le 
\sum_{n \ge 0} \frac{\|A^n\|\|x\|}{(\rho +\epsilon)^n} \le  
\sum_{n \ge 0} \frac{C(\rho + \epsilon/2)^n}{(\rho +\epsilon)^n} \|x\| = 
M \|x\|$$ 
for some constant $M$. Therefore, $N$ is a norm in the underlying space, equivalent to the norm $\| \|$.  
Note that $N(Ax) = \sum_{n \ge 0} \frac{\|A^{n+1}x\|}{(\rho +\epsilon)^n} =
(\rho +\epsilon) \ \sum_{n \ge 0} \frac{\|A^{n+1}x\|}{(\rho +\epsilon)^{n+1}} \le 
(\rho +\epsilon) \ \sum_{n \ge 0} \frac{\|A^{n}x\|}{(\rho +\epsilon)^{n}} =
(\rho +\epsilon) \ N(x)$, i.e.: the norm of $A$ in the operator norm derived from $N$ satisfies 
$ N(A) \le \rho + \epsilon $. 
