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

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

Yiorgos S. Smyrlis
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afsdf dfsaf
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  • The title suggests you're looking for a matrix $D$ such that the equality in the title holds, but this doesn't seem to be like what you want. Even if you fix this, the title only deals with half of the question. Please find a better title. Edit: This was true before Yiorgos S. Smyrlis partially fixed the title for you. – Git Gud Jan 26 '14 at 22:41
  • Note that your breakdowns rely essentially on the fact that $A$ and $B$ (or alternately, $E$ and $F$) commute; without this, you cannot use $e^{A+B}=e^Ae^B$. Also, that isn't even what you've written for $e^C$; instead, you seem to be using the completely incorrect identity $e^{A+B} = e^A+e^B$. – Steven Stadnicki Aug 13 '14 at 21:47

3 Answers3

5

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

Git Gud
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5

$D=2I+F$, 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.

Yiorgos S. Smyrlis
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  • how did you come up with $$ \exp(F)=\left(\begin{matrix}\cos 1&-\sin 1\ \sin 1&\cos 1\end{matrix}\right), $$ is there any intermediate step in between? – afsdf dfsaf Jan 26 '14 at 23:06
  • I hope it's not too much. I added the details that OP asked for and fixed a typo. @Yiorgos S. Smyrlis – Patrick Da Silva Jan 26 '14 at 23:11
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$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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 $\pars{1}$: $$\color{#00f}{\large% \expo{D} = \pars{% \begin{array}{rr} \expo{2}\cos\pars{1} & -\expo{2}\sin\pars{1} \\[1mm] \expo{2}\sin\pars{1} & \expo{2}\cos\pars{1} \end{array}}} $$
Felix Marin
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  • it should be $$\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}}} $$? – afsdf dfsaf Jan 27 '14 at 03:18
  • @afsdfdfsaf Yes. I agree. I already check my answer. Thanks. – Felix Marin Jan 27 '14 at 04:39