Let $A$ be a complex $n\times n$ matrix where $n \geq 2$ such that $A$ is not normal. Then I want to show that the unital, commutative Banach algebra generated by $A$, i.e. $\{p(A): p \text{ polynomial} \}$, has non-trivial radical, i.e. the intersection over all maximal ideals is not equal to $\{0\}$.
My thoughts: Assume $A$ were normal, then we may consider the generated unital commutative C*-algebra. Of course this is semisimple since all C*-algebras are. However we may get some further insight: This algebra is isomorphic to continuous functions on its spectrum. The maximal ideals there are functions vanashing in one point. For the matrix algebra this means that the maximal ideals are the subalgebras of matrices which vanish on a fixed eigenvector. As $n\geq 0$, we have two distinct eigenvectors and the claim follows. I think this is the geometric reason for my original claim to be true, but I have no idea how to do it.
It is not true. Normality is tied to a particular inner product, whereas the algebraic structure of the unital algebra generated by $A$ is not necessarily. If $A$ is diagonalizable, the algebra generated by $A$ is isomorphic to the algebra generated by a diagonal matrix with the same set of eigenvalues. If $D$ is a diagonal matrix with distinct eigenvalues, then the algebra generated by $D$ is the algebra of all diagonal matrices, which is semisimple.
For a concrete example, with the standard inner product, $A=\begin{bmatrix}1&1\\0&0\end{bmatrix}$ is not normal, but the unital algebra generated by $A$ is $\left\{aI+b A=\begin{bmatrix}a+b&b\\0&a\end{bmatrix}:a,b\in\mathbb C\right\}$, which has maximal ideals $\left\{\begin{bmatrix}c&c\\0&0\end{bmatrix}:c\in\mathbb C\right\}$ and $\left\{\begin{bmatrix}0&c\\0&-c\end{bmatrix}:c\in\mathbb C\right\}$, with intersection $\{0\}$.