= Solution
A norm is positive definite, homogeneous, and subadditive; equivalence means mutual bounds by positive constants. Finite-dimensional norm equivalence gives corresponding constants for operator norms, whose $k$th roots tend to one, so $\rho(A)$ is norm-independent. If $\|A\|\lt 1$, Picard iteration gives $(I-A)^{-1}=\sum_{k\ge0}A^k$. If $\rho(A)\lt 1$, choose $\rho(A)\lt \alpha\lt 1$ and define $\|v\|_\alpha=\sup_{m\ge0}\alpha^{-m}\|A^mv\|$; it is finite, equivalent to the original norm, and $\|Av\|_\alpha\le\alpha\|v\|_\alpha$.
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