symmetric. In this case $\|A\|=\rho(A)$, $\|\hat A\|_2 = \rho (\hat A)$ and item 2 reduces to item 1. On the other hand, in the more general case where $A$ is normal, the matrix $\hat A$ is not necessarily symmetric or even normal and we obtain the bound $$\|A\| = \rho(A)\le \rho(\hat A)\,,$$ which is generally better than $\|A\|\le \|\hat A\|_2$.

Next, we prove a simple invertibility condition for matrices over the commutative $C^*$-algebra $C(\mathcal{X})$.

\[thm\_invert\] A matrix $A$ over $\mathcal{A}_0 = C(\mathcal{X})$ is invertible in $\mathcal{A}=\mathcal{M}_n(\mathcal{A}_0)$ if and only if for each $t\in \mathcal{X}$ the $n\times n$ matrix $A(t) = [a_{ij}(t)]$ is invertible in $\mathcal{M}_n(\mathbb{C})$.

The theorem will be proved if we show that $$\label{spectra}
\sigma (A) = {\bigcup}_{t\in \mathcal{X}}\sigma (A(t))\,.$$ Assume first that $\lambda\in \sigma(A(t_0))$ for some $t_0\in \mathcal{X}$. Then $\lambda I_n - A(t_0)$ is not invertible, therefore $\lambda I_n - A(t)$ is not invertible for all $t\in \mathcal{X}$, and $\lambda I_n - A$ fails to be invertible as an element of $\mathcal{A}=\mathcal{M}_n(\mathcal{A}_0)$. This shows that $${\bigcup}_{t\in \mathcal{X}}\sigma (A(t))\subseteq \sigma (A).$$ To prove the reverse inclusion, we show that the resolvent sets satisfy $${\bigcap}_{t\in \mathcal{X}}r(A(t)) \subseteq r(A).$$ Indeed, if $z\in r(A(t))$ for all $t\in \mathcal{X}$, then the matrix-valued function 