Why is $\rho(A) = \lim_{n \to \infty}\|A^n\|^{1/n}$ for any matrix norm $\|\cdot\|$?
My professor told me this but I cannot find the proof of it. It is rather trivial for norms like $\|\cdot\|_2$ but the statement holds true for all kinds of matrix norms, even including the inconsistent ones. So right now I have no clue how to prove this yet. Any thoughts will be appreciated.
Edit: Is this derived by equivalence of norm on euclidean space and the fact that for every $\epsilon > 0$ and a given $A$, one can find matrix norm $\|\cdot\|_\epsilon$ such that $\|A\|_\epsilon \leq \rho(A) + \epsilon$?
This can be found in various places, such as Wikipedia. This even holds for an arbitrary Banach algebra, which can be found in any book on functional analysis that covers the topic of Banach algebra. The key ingredient toward the general result is the Fekete's lemma.
However, in your particular case, if you admit this holds for $\|\cdot\|_2$, then it's easy to see why it holds for any other norm. Since $n\times n$ matrices form a finite dimensional space, all norms are equivalent, and say $\|\cdot\|\le C\|\cdot\|_2$, we have $\|A^n\|\le C\|A^n\|_2$, $\|A^n\|^{1/n}\le C^{1/n}\|A^n\|^{1/n}_2$, and let $n\rightarrow\infty$, we get $$\limsup_{n\rightarrow\infty} \|A^n\|^{1/n}\le\lim_{n\rightarrow\infty} \|A^n\|_2^{1/n}$$
By symmetry, we can also establish that $\lim_{n\rightarrow\infty} \|A^n\|_2^{1/n}\le\liminf_{n\rightarrow\infty} \|A^n\|^{1/n}$, hence we know that not only $\lim_{n\rightarrow\infty} \|A^n\|^{1/n}$ exists but equals $\lim_{n\rightarrow\infty} \|A^n\|_2^{1/n}$.