Gelfand's spectral radius formula

Step 1: $\rho(A)\le\|A^k\|^{1/k}$ for every $k$. If $Av=\lambda v$ with $v\ne0$ ($v$ complex) and $|\lambda|=\rho(A)$, then $A^kv=\lambda^kv$, so $\rho(A)^k=\rho(A^k)$ (spectral mapping with $p(x)=x^k$). For a *submultiplicative matrix norm induced by a vector norm*, $\|A^k\|\ge\rho(A^k)$ since $|A^kv|=|\lambda|^k|v|$ gives $\|A^k\|\ge|\lambda|^k$. For a general submultiplicative norm that is not induced by a vector norm, use the equivalence of norms: $\|\cdot\|\ge c\,\|\cdot\|_{op}$ for some $c>0$, so $\|A^k\|^{1/k}\ge c^{1/k}\rho(A)$, and $c^{1/k}\to1$. Either way,
$$\liminf_{k}\|A^k\|^{1/k}\ge\rho(A).$$

Step 2: $\limsup\le\rho(A)$. Fix $\varepsilon>0$ and set $B=A/(\rho(A)+\varepsilon)$. Then $\rho(B)=\frac{\rho(A)}{\rho(A)+\varepsilon}<1$, so $B^k\to0$ (previous problem), hence $\|B^k\|\le1$ for all large $k$. Equivalently
$$\|A^k\|\le(\rho(A)+\varepsilon)^k\quad\text{for large }k,\qquad\text{so}\qquad\|A^k\|^{1/k}\le\rho(A)+\varepsilon.$$
Hence $\limsup\|A^k\|^{1/k}\le\rho(A)+\varepsilon$ for every $\varepsilon>0$.

Conclusion. $\liminf\ge\rho(A)\ge\limsup-\varepsilon$ for all $\varepsilon$; the limit exists and equals $\rho(A)$.

Example. $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ has $\rho=0$ and $A^2=0$: $\|A^k\|^{1/k}=0$ for $k\ge2$. For $A=\begin{pmatrix}0.9&1\\0&0.9\end{pmatrix}$, $\|A^{50}\|^{1/50}\approx0.9\cdot1.0\!\ldots$ approaches $0.9$ slowly.

Remark. As a corollary, $\rho(A)=\inf_{\|\cdot\|}\|A\|$ over all operator norms: for each $\varepsilon$ there is a norm with $\|A\|<\rho(A)+\varepsilon$. This is the topological meaning of the spectral radius.