Let $z \in \mathbb{C}$.
Let $ f(z)= \begin{cases} 1,& \text{if } |z|<\frac{1}{2}\\ 0, & \text{if} |z-2| <\frac{1}{2} \\ 0, & \text{if} |z-3| <\frac{1}{2} \end{cases}$
Let \begin{equation*} A = \begin{pmatrix} 4 & 1 & 0 & -4 & 0 & 0\\ 0 & 4 & -1 & 0 & -4 & 2\\ 0 & 0 & 6 & 0 & 0 & -6\\ 2 & 0 & 0 & -2 & 1 & 0\\ 0 & 2 & -1 & 0 & -2 & 2\\ 0 & 0 & 3 & 0 & 0 & -3\\ \end{pmatrix}\end{equation*}
Find $f(A)$.
Attempt:
We know that $\mathbb{C}^6$ is a Hilbert space. and $B(\mathbb{C}^6) \simeq \mathbb{M_6}$. So we can consider $A$ as a function in $B(\mathbb{C}^6)$
And we can write $Ae_j= \sum_{n=1}^6a_n e_n$, where $(e_n)$ is the orthonormal basis for $\mathbb{C}^6$.
Also I know that the spectrum $\sigma(A)= \{\text{eigenvalues of} A \}$
From here I'm not sure how to proceed.
I'm trying to solve this using Banach Algebra point of view. But any other suggestions will be much appreciated too!
EDIT: My apologies, there was a typo in the original question. Now corrected
Thank you!
Important edit: I just realized that $A$ is not a normal matrix. The technique below would work in general for a normal matrix, however I am not sure how $f(A)$ is even defined for a non-normal matrix. The continuous (or Borel for von Neumann algebras) functional calculus makes sense for normal elements only, as far as I remember.
Original Post:
Plugging the matrix in a calculator, you get the eigenvalues: $\sigma(A)=\{0,2,3\}$. The function $f$ you are defining is not defined on $0$, so I will write $c=f(0)$. If you can find a polynomial $p(z)$ such that $p(0)=c$, $p(2)=f(2)=0$ and $p(3)=f(3)=0$, (for example $p(z)=\frac{c}{6}(z-2)(z-3)$ is such a polynomial) then since $f=p$ on $\sigma(A)$ we have that $f(A)=p(A)$. But $p(A)$ is easily computed, for example if $p$ is the polynomial I mentioned earlier then $$p(A)=\frac{c}{6}(A-2I)(A-3I).$$