 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 $\Vert A\Vert <1$, Picard iteration gives $(I-A{)}^{-1}={\sum }_{k\ge 0}{A}^k$. If $\rho (A)<1$, choose $\rho (A)<\alpha <1$ and define $\Vert v{\Vert }_{\alpha }={\sup }_{m\ge 0}{\alpha }^{-m}\Vert A^{m}v\Vert$; it is finite, equivalent to the original norm, and $\Vert Av{\Vert }_{\alpha }\le \alpha \Vert v{\Vert }_{\alpha }$.