Find $\exp(D)$ where $D = \begin{bmatrix}2& -1 \\ 1 & 2\end{bmatrix}. $ $$C = \begin{bmatrix}2& -1 \\ 0 & 2\end{bmatrix}\quad   $$
I break it down into two matrices
$$A = \begin{bmatrix}2& 0 \\ 0 & 2\end{bmatrix}\quad \text{and}\quad 
B =\begin{bmatrix}0 & -1 \\ 0 & 0\end{bmatrix}.$$
For matrix $A$, $$\operatorname{exp}(A) = \begin{bmatrix}e^2& 0 \\ 0 & e^2\end{bmatrix}\quad.$$
For matrix $B$, we have that
$B^k=0$, for all $k\ge 2$$$
\exp B=I +B+\frac{B^2}{2!}+\cdots+\frac{B^n}{n!}+\cdots=
\cdots=I+B 
=\begin{bmatrix}1 & -1 \\ 0 & 1\end{bmatrix}.$$
So exp(C) = $$ \begin{bmatrix} \mathrm{e^2}+1 & -1\\ 0 &  \mathrm{e^2}+1\end{bmatrix}\quad   $$
Can someone check to see if this is right?
if so my next question is to find
$$D = \begin{bmatrix}2& -1 \\ 1 & 2\end{bmatrix}\quad   $$
I break it down into two matrices
$$E = \begin{bmatrix}2& 0 \\ 0 & 2\end{bmatrix}\quad \text{and}\quad 
F =\begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}$$
For matrix $E$, $\exp(E)$ is the same as $\exp(A)$,
but for matrix $F$, I cannot apply the same method to solve matrix $B$. I am wondering how to find the $\exp(F)$?
Thank you
 A: Hint: $D{}$ is diagonalizable (over $\mathbb C$).
In the first part you are apparently using the false rule $\exp(X+Y)=\exp(X)+\exp(Y)$. this is false in general and what you did is wrong. Note however that $AB=BA$, therefore $\exp(A+B)=\exp(A)\exp(B)$.
A: $D=2I+F$, and thus (as $I$ and $F$ commute)
$$
\exp(D)=\exp(2I)\exp(F)=\mathrm{e}^2\exp(F).
$$
But
$$
F^2=-I,\,\,F^3=-F,\,\,F^4=I,\,\,F^5=F, \mathrm{etc}. 
$$
Hence since $\cos x = \sum_{k \ge 0} \frac{x^{2k}}{(2k)!}$ and $\sin(x) = \sum_{k \ge 0} \frac{x^{2k+1}}{(2k+1)!}$, we have
$$
\exp(F)=\left(\begin{matrix}\cos 1&-\sin 1\\ \sin 1&\cos 1\end{matrix}\right),
$$
and finally
$$
\exp(D)=\mathrm{e}^2\left(\begin{matrix}\cos 1&-\sin 1\\ \sin 1&\cos 1\end{matrix}\right).
$$
Note that
$$
\cos x = \frac{e^{ix} + e^{-ix}}2, \quad \sin x = \frac{e^{ix} - e^{-ix}}{2i},
$$
hence you can simplify your answer by plugging $i$ into these expressions and substituing in the matrices.
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\dd}{{\rm d}}%
 \newcommand{\down}{\downarrow}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
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\begin{align}&
D = \pars{\begin{array}{rr}2 & -1\\ 1 & 2\end{array}} =
2\pars{\begin{array}{cc}1 & 0\\ 0 & 1\end{array}} - \ic\
\overbrace{\pars{\begin{array}{rr}0 & -\ic\\ \ic & 0\end{array}}}
^{\ds{ \sigma_{y}}}
\\[3mm]&\quad\imp\quad
D = 2 -\ic\sigma_{y}\quad\imp\quad
{\large\expo{D} = \expo{2}\expo{-\ic\sigma_{y}}}\tag{1} 
\end{align}
where $\ds{\sigma_{y}}$ is a Pauli Matrix. Notice that $\ds{\sigma_{y}^{2} = 1}$.
Let's consider $\expo{-\ic\mu\sigma_{y}}$. It satisfies
$\ds{\pars{\totald[2]{}{\mu} + 1}\expo{-\ic\mu\sigma_{y}} = 0}$ with
$\ds{\left.\expo{-\ic\mu\sigma_{y}}\right\vert_{\mu = 0} = 1}$
and
$\ds{\left.\totald{\expo{-\ic\mu\sigma_{y}}}{\mu}\right\vert_{\mu = 0} = -\ic\sigma_{y}}$.
$$
\mbox{It leads to}\
\expo{-\ic\mu\sigma_{y}} = \cos\pars{\mu} - \ic\sigma_{y}\sin\pars{\mu}
\quad\imp\quad
\expo{-\ic\sigma_{y}}
=
\pars{%
\begin{array}{rr}
\cos\pars{1} & -\sin\pars{1}
\\
\sin\pars{1} & \cos\pars{1}
\end{array}}
$$

With result $\pars{1}$:
$$\color{#00f}{\large%
\expo{D} =
\pars{%
\begin{array}{rr}
\expo{2}\cos\pars{1} & -\expo{2}\sin\pars{1}
\\[1mm]
\expo{2}\sin\pars{1} & \expo{2}\cos\pars{1}
\end{array}}}
$$

