I want to show that for diagonalizable matrices $A,B$ that the spectral radius (defined as $$\rho(A) = \lim_{k\to\infty}\|A^k\|^{1/k} = \max_{1\le i \le n} |\lambda_i|$$gives us $$\rho(AB) \le \rho(A)\rho(B).$$
Immediately I thought for any matrix norm, $$\begin{align}\|(AB)^n\| & = \|B^nA^n\|\\ &\le \|B^n\|\|A^n\|\\&=\|A^n\|\|B^n\|\\\implies \|(AB)^n\|^{1/n} &\le\|A^n\|^{1/n}\|B^n\|^{1/n} \\ \implies\lim_{n\to\infty}\|(AB)^n\|^{1/n}&\le\lim_{n\to\infty}\|A^n\|^{1/n}\lim_{n\to\infty}\|B^n\|^{1/n} \\ \implies\rho(AB) &\le \rho(A)\rho(B).\end{align}$$
However, it occurred to me shortly after that this isn't necessarily true, that $(AB)^n \ne B^nA^n$ the same way that $(AB)^{-1} = B^{-1}A^{-1}$ and $(AB)^T = B^TA^T$. This is only true if $AB = BA$.
Never anywhere did I use the definition of diagonalizable for either matrices $A$ or $B$.
Most sources I can find say that $\rho(AB)\le\rho(A)\rho(B)$ if and only if $AB = BA$, which leads me to believe there might actually be a condition missing on this problem. Moreover, I simply need to show that there exists a matrix norm $\|\cdot\|$ over $M_n(\mathbb C)$ such that $\|A\| = \rho(A)$ for a diagonalizable matrix $A$. This is just one of the intermediate steps of showing it's a matrix norm. But I'm unfortunately stuck.
That's not true. $\rho(AB)$ can be greater than $\rho(A)\rho(B)$ when $A$ and $B$ are diagonalisable. E.g. consider $$ A=\pmatrix{1&2\\ 2&1}, \ B=\pmatrix{1&1\\ 2&1}, \ AB=\pmatrix{5&3\\ 4&3}. $$ Their spectra are respectively $\{3,-1\}, \{1+\sqrt{2},1-\sqrt{2}\}$ and $\{4+\sqrt{13},4-\sqrt{13}\}$ but $\rho(AB)=4+\sqrt{13}\approx 7.61>7.24\approx 3(1+\sqrt{2})=\rho(A)\rho(B)$.
And you said that
While the backward implication (the "if" part) is obviously true when both $A$ and $B$ are diagonalisable, the forward implication is in general false. It is easy to construct a counterexample, e.g. $A=\pmatrix{1&0\\ 0&0}$ and $B=\pmatrix{1&1\\ 1&1}$.