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'm very confused about how to approach this problem.
It would be nice if someone could walk me through how to solve this. But any hints or ideas will be appreciated!
Thank you!
$$C=\begin{pmatrix} 2B-A & 2A-2B\\ B-A & 2A-B \end{pmatrix}$$
Let $U=\begin{pmatrix} -1 & 2\\ -1 & 2 \end{pmatrix}$ and $V=\begin{pmatrix}2 & -2\\ 1 & -1 \end{pmatrix}$.
Then $C=U\otimes A+V\otimes B$, $U^2=U$, $V^2=V$ and $UV=VU=0$.
Hence, using the Kronecker mixed product,
$$C^2=(U\otimes A)^2+(V\otimes B)^2+(U\otimes A)\cdot(V\otimes B)+(V\otimes B)\cdot(U\otimes A)$$
$$C^2=U^2\otimes A^2+V^2\otimes B^2+(UV)\otimes (AB)+(VU)\otimes(BA)$$
$$C^2=U\otimes A^2+V\otimes B^2$$
By induction, we would find that
$$C^n=U\otimes A^n+V\otimes B^n$$
Note that the preceding works also for $C^0$, because $U+V=I_2$, hence $$U\otimes A^0+V\otimes B^0=U\otimes I_3+V\otimes I_3=(U+V)\otimes I_3=I_6$$
Therefore
$$\exp C=U\otimes \exp A+V\otimes \exp B$$
$A$ is nilpotent and $A^3=0$ hence
$$\exp A=I+A+\frac12A^2=\begin{pmatrix}1 & 1 & \frac12\\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}$$
$B$ is a Jordan matrix, and the first Jordan block is $\begin{pmatrix}2 & 1 \\ 0 & 2\end{pmatrix}=2\begin{pmatrix}1 & \frac12 \\ 0 & 1\end{pmatrix}$.
And $\begin{pmatrix}1 & \frac12 \\ 0 & 1\end{pmatrix}^n=\begin{pmatrix}1 & \frac n2 \\ 0 & 1\end{pmatrix}$, hence
$$B^n=\begin{pmatrix}2^n & n 2^{n-1} & 0\\ 0 & 2^n & 0 \\ 0 & 0 & 3^n\end{pmatrix}$$
And
$$\sum_{n=0}^{\infty} \frac{n2^{n-1}}{n!}=\sum_{n=1}^{\infty} \frac{2^{n-1}}{(n-1)!}=e^2$$
Hence
$$\exp B=\begin{pmatrix} e^2 & e^2 & 0 \\ 0 & e^2 & 0 \\ 0 & 0 & e^3\end{pmatrix}$$
All in all,
$$\exp C=\begin{pmatrix} -1 & 2\\ -1 & 2 \end{pmatrix}\otimes\begin{pmatrix}1 & 1 & \frac12\\ 0 & 1 & 1 \\ 0 & 0 & 1\end{pmatrix}+ \begin{pmatrix}2 & -2\\ 1 & -1 \end{pmatrix}\otimes\begin{pmatrix} e^2 & e^2 & 0 \\ 0 & e^2 & 0 \\ 0 & 0 & e^3\end{pmatrix}$$
$$\exp C= \begin{pmatrix} 2e^2-1 & 2e^2-1 & -\frac12 & -2e^2+2 & -2e^2+2 & 1\\ 0 & 2e^2-1 & -1 & 0 & -2e^2+2 & 2\\ 0 & 0 & 2e^3-1 & 0 & 0 & -2e^3+2\\ e^2-1 & e^2-1 & -\frac12 & -e^2+2 & -e^2+2 & 1\\ 0 & e^2-1 & -1 & 0 & -e^2+2 & 2 \\ 0 & 0 & e^3-1 & 0 & 0 & -e^3+2\end{pmatrix}$$
And the hint with $D=\begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix}$?
Well, we have already noticed that $U$ and $V$ are two commuting projectors, and we have additionally:
$$D^{-1}UD=\begin{pmatrix} 0 & 0 \\ 0 & 1\end{pmatrix}$$ $$D^{-1}VD=\begin{pmatrix} 1 & 0 \\ 0 & 0\end{pmatrix}$$
That is, $U$ projects on the subspace spanned by $(1, 1)^T$, and $V$ projects on the subspace spanned by $(2, 1)^T$.
The eigendecomposition might have helped to discover that $U^2=U$, $V^2=V$ and $UV=VU=0$, which are needed to simplify $C^n$.