$t\mapsto (zI_n - A(t))^{-1}$ is well defined and necessarily continuous on $\mathcal{X}$. Hence, $z\in r(A)$. This completes the proof.

Clearly, the set $\mathcal{K} = \{A(t)\,;\, t\in \mathcal{X}\}$ is compact in $\mathcal{M}_n (\mathbb{C})$. The spectral radius, as a function on $\mathcal{M}_n (\mathbb{C})$, is continuous and thus the function $t\mapsto \rho(A(t))$ is continuous on the compact set $\mathcal{X}$. It follows that this function attains its maximum value for some $t _0\in \mathcal{X}$. The equality (\[spectra\]) above then implies that $$\rho(A) = \rho(A(t_0)) = \max_{t\in \mathcal{X}} \rho(A(t))\,.$$ If $A$ is normal, we obviously have $$\label{norms}
\|A\| = \max_{t\in \mathcal{X}} \|A(t)\|_2\,.$$ Recalling that for any $A\in \mathcal{A}$ the norm satisfies $\|A\| = \sqrt{\rho(AA^*)} = \sqrt{\rho(A^*A)}$, we conclude that the identity (\[norms\]) holds for any matrix $A$ over the $C^*$-algebra $C(\mathcal{X})$.

As a special case, consider an $n\times n$ matrix with entries in $C(\mathcal{X})$, where $\mathcal{X}=[0,1]$. Each entry $a_{ij} = a_{ij}(t)$ of $A$ is a continuous complex-valued function of $t$. One can think of such an $A$ in different ways. As a mapping of $[0,1]$ into $\mathcal{M}_n(\mathbb{C})$, $A$ can be regarded as a continuous curve in the space of all $n\times n$ complex matrices. On the other hand, $A$ is also a point in the 