I have to do the next problem but I'm a little lost.
Let $A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix}$ be two elements of $\mathcal{M}_3(\mathbb{C})$.
Let $$ T = \begin{bmatrix} 4 & 1 & 0 & -4 & 0 & 0 \\ 0 & 4 & -1 & 0 & -4 & 2 \\ 0 & 0 & 6 & 0 & 0 & -6 \\ 2 & 0 & 0 & -2 & 1 & 0 \\ 0 & 2 & -1 & 0 & -2 & 2 \\ 0 & 0 & 3 & 0 & 0 & -3 \end{bmatrix} \in \mathcal{M}_6(\mathbb{C}). $$
a) Let $\exp(z) = e^z$, $z \in \mathbb{C}$. Find $\exp(T)$.
Hint: $$S := \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix}$$
b) Let $g(z) = \begin{cases} 1 & \mbox{ if } | z | < \frac{1}{2} \\ 0 & \mbox{ if } |z - 2 | < \frac{1}{2} \\ 0 & \mbox{ if } |z-3| < \frac{1}{2} \end{cases}$. \ \ \ \ \ Find $g(T)$.
I was able to do a) I use the hint to obtain this equation
$$ \begin{bmatrix} 1 & 2 \\ 1 & 1 \end{bmatrix} \ \begin{bmatrix} A & 0 \\ 0 & B \end{bmatrix}\ \begin{bmatrix} -1 & 2 \\ 1 & -1 \end{bmatrix} = \begin{bmatrix} -A + 2 B & 2 A - 2 B \\ -A + B & 2 A - B \end{bmatrix}=T$$
and then $exp(z)=\lim_{n\to\infty}\sum_{i=0}^n\frac{z^i}{i!}$.
I need help for b)
What I though was of using Functional calculus then if $\Gamma$ is a closed countour of $\sigma(T)$ then
$$g(T)=\int_\Gamma g(z)(z-T)^{-1} dz$$
I though of defining $\Gamma$ as the union of $|z|=1/2, |z-2|=1/2,|z-3|=1/2$. (I still have to prove that $\sigma(T)$ is in the domain of $g$). My problem is that I'm not sure if this is a correct aproach, if there is a easy way to prove that $\sigma(T)$ is contain in the domaing of $g$ and lastly I'm not sure how to aproach the integral. How should I calculated.