Find $\exp(D)$ where $D = \begin{bmatrix}2& -1 \\ 1 & 2\end{bmatrix}. $

Question

$$C = \begin{bmatrix}2& -1 \\ 0 & 2\end{bmatrix}\quad $$

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

Answer 1

$D=2I+F$, 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.

Answer 2

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 $\exp(A+B)=\exp(A)\exp(B)$.

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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 $\pars{1}$: $$\color{#00f}{\large% \expo{D} = \pars{% \begin{array}{rr} \expo{2}\cos\pars{1} & -\expo{2}\sin\pars{1} \\[1mm] \expo{2}\sin\pars{1} & \expo{2}\cos\pars{1} \end{array}}} $$