find the matrix exponential Let
\begin{equation*}
A = 
\begin{pmatrix}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{pmatrix}\end{equation*}
\begin{equation*}
B = 
\begin{pmatrix}
2 & 1 & 0\\
0 & 2 & 0\\
0 & 0 & 3
\end{pmatrix}\end{equation*}
\begin{equation*}
C = 
\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*}
Suppose $\exp(z)=e^z$, for $z\in \mathbb{C}$. Find $\exp(C)$.
Hint:
\begin{equation*}
D = 
\begin{pmatrix}
2 & 1 \\
1 & 1 \end{pmatrix}\end{equation*}
is invertible.
I realize that this is already asked. But I'm hoping to get a way of solving this problem in the  Banach algebra prospective.
Attempt:
We know that $\mathbb{C}^n$ is a Hilbert space. and $B(\mathbb{C}^n) \simeq \mathbb{M_n}$. So we can consider each matrix as a function in $B(\mathbb{C}^n)$
And we can write $Ae_j= \sum_{k=1}^n a_k e_k$, where $(e_k)$ is the orthonormal basis for $\mathbb{C}^n$.
Also I know that the spectrum $\sigma(A)= \{\text{eigenvalues of} A \}$
Since the $\mathcal{M_n}$ is over $\mathcal{C}$, it is a matrix ring.
By the hint, I know that
\begin{equation*}
D^{-1}= 
\begin{pmatrix}
1 & -1 \\
-1 & 2 \end{pmatrix}\end{equation*}
And we have $D\begin{pmatrix}
x & 0 \\
0 & y \end{pmatrix}D^{-1}= 
\begin{pmatrix}
-x+2y & 2x-2y \\
-x+y & 2x-y \end{pmatrix}$.
From here, I wonder if we could perform some sort of induction from $B(\mathbb{C}^2) \simeq \mathbb{M_2}$ to $B(\mathbb{C}^6) \simeq\mathbb{M_6}$.
Any suggestions will be much appreciated.
 A: Rye, since we discussed this a little bit in your previous question: Note that for the case $f=\exp$ there is a standard way of defining $f(a)$ for non-normal elements of a unital Banach algebra:
$$f(a)=\sum_{k=0}^\infty\frac{a^k}{k!}$$
You can find more details about this in the first pages of Murphy's book. This can also be done for any entire function: if $g(z)=\sum_{k=0}^\infty a_kz^k$, then set $g(x)=\sum_{k=0}^\infty a_kx^k$ for $x\in A$. The series converges in the Banach algebra since it converges absolutely and we are in a Banach space. Anyway, back to your question:
We can use block-form: set
Note that $$C=\begin{pmatrix}-A+2B& 2A-2B\\ -A+B&2A-B\end{pmatrix}$$
If $P=\begin{pmatrix}B&0\\0&A\end{pmatrix}$ and $$E=\begin{pmatrix}2I_3&I_3\\I_3&I_3\end{pmatrix}$$
then $E$ is invertible, $E^{-1}=\begin{pmatrix}I_3&-I_3\\-I_3&2I_3\end{pmatrix}$ and $C=EPE^{-1}$. From this we have that $C^n=EP^nE^{-1}$ and from the definition of $\exp(\cdot)$ we will have that $\exp(C)=E\cdot\exp(P)\cdot E^{-1}$.
It is easily verified that $\exp(P)=\begin{pmatrix}\exp(B)&0\\0&\exp(A)\end{pmatrix}$, so the whole problem has been reduced to computing the exponentials of $A$ and $B$ (which are in $M_3(\mathbb{C})$, life is much easier now).
But $A^3=0$, so $\exp(A)=I_3+A+\frac{1}{2}A^2$. On the other hand, for $n\geq1$
$$B^n=\begin{pmatrix}2^n&n\cdot 2^{n-1}&0\\0&2^n&0\\0&0&3^n\end{pmatrix} $$
can be easily verified (use induction). Observe that
$$\sum_{n=0}^\infty\frac{n2^{n-1}}{n!}=\sum_{n=1}^\infty\frac{2^{n-1}}{(n-1)!}=\sum_{n=0}^{\infty} \frac{2^n}{n!}=e^2 $$
so
$$\exp(B)=\begin{pmatrix}e^2&e^2&0\\0&e^2&0\\0&0&e^3\end{pmatrix} $$
Now that we know $\exp(A)$, $\exp(B)$ and $E$, we plug everything back to
$$\exp(C)=E\exp(P)E^{-1}$$
and the result follows.
