Problem calculating the sine of a matrix

Question

Given the matrix $A=\begin{pmatrix}-\frac{3\pi}{4} & \frac{\pi}{2}\\\frac{\pi}{2}&0\end{pmatrix}$, I want to calculate the sine $\sin(A)$.

I do so by diagonalizing A and plugging it in the power series of the sine:

\begin{align} \sin (A) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} A^{2k+1}. \end{align}

The diagonalization leads to:

\begin{align} A = \frac{1}{5} \begin{pmatrix}-2 & 1\\1&2\end{pmatrix} \begin{pmatrix}-\pi & 0\\0&\frac{\pi}{4}\end{pmatrix} \begin{pmatrix}-2 & 1\\1&2\end{pmatrix} \end{align} and thus: \begin{align} A^n = \frac{1}{5} \begin{pmatrix}-2 & 1\\1&2\end{pmatrix} \begin{pmatrix}-\pi & 0\\0&\frac{\pi}{4}\end{pmatrix}^n \begin{pmatrix}-2 & 1\\1&2\end{pmatrix}. \end{align} Hence: \begin{align} \sin (A) &= \begin{pmatrix}-2 & 1\\1&2\end{pmatrix} \begin{pmatrix}\sin(-\pi) & 0\\0&\sin(\frac{\pi}{4})\end{pmatrix} \begin{pmatrix}-2 & 1\\1&2\end{pmatrix}\\ &= \begin{pmatrix}-2 & 1\\1&2\end{pmatrix} \begin{pmatrix}0 & 0\\0&\frac{1}{\sqrt{2}})\end{pmatrix} \begin{pmatrix}-2 & 1\\1&2\end{pmatrix}\\ &= \frac{1}{5}\begin{pmatrix}\frac{1}{\sqrt{2}} & \sqrt{2}\\\sqrt{2}&2\sqrt{2}\end{pmatrix}, \end{align} which differs from "Wolfram Alpha's result" \begin{align} \sin(A) &= \begin{pmatrix}-\frac{1}{\sqrt{2}} & 1\\ 1 & 0 \end{pmatrix} . \end{align}

How can this happen?

Answer 1

The problem is that Wolfram Alpha interprets "sin(A)" for a matrix A (or array of however many dimensions, or list of list of lists, or what have you) as meaning simply the result of applying sin component-wise.

This is not what you intended, and you did your intended calculation perfectly fine.

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\,\mathsf{A} \equiv \pars{\begin{array}{rc} \ds{-\,{3\pi \over 4}} & \ds{\pi \over 2} \\ \ds{\pi \over 2} & \ds{0} \end{array}}\,,\qquad\sin\pars{\mathsf{A}} =\, ?}$

$$ \mbox{Note that}\quad\,\mathsf{A} = -\,{3\pi \over 8}\,\sigma_{0} + {\pi \over 2}\,\sigma_{x} - {3\pi \over 8}\,\sigma_{z} = -\,{3\pi \over 8}\,\sigma_{0} + \vec{b}\cdot\vec{\sigma}\,,\qquad \left\lbrace\begin{array}{rcr} \ds{b_{x}} & \ds{=} & \ds{\pi \over 2} \\[1mm] \ds{b_{y}} & \ds{=} & \ds{0} \\[1mm] \ds{b_{z}} & \ds{=} & \ds{-\,{3\pi \over 8}} \end{array}\right. $$ where $\ds{\sigma_{0}}$ is the $\ds{2 \times 2}$ identity matrix. $\ds{\braces{\sigma_{i},\ i = x,y,z}}$ are the $\ds{2 \times 2}$ Pauli Matrices which satisfies $$ \sigma_{i}^{2} = \sigma_{0}\,,\qquad \left\lbrace\begin{array}{rcccl} \ds{\sigma_{x}\sigma_{y}} & \ds{=} & \ds{-\sigma_{y}\sigma_{x}} & \ds{=} & \ds{\ic\sigma_{z}} \\ \ds{\sigma_{y}\sigma_{z}} & \ds{=} & \ds{-\sigma_{z}\sigma_{y}} & \ds{=} & \ds{\ic\sigma_{x}} \\\ds{\sigma_{z}\sigma_{x}} & \ds{=} & \ds{-\sigma_{x}\sigma_{z}} & \ds{=} & \ds{\ic\sigma_{y}} \end{array}\right. $$

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$\ds{\expo{\mu\vec{b}\cdot\vec{\sigma}}}$ satisfies $\ds{\pars{\partiald[2]{}{\mu} - \vec{b}\cdot\vec{b}}\expo{\mu\vec{b}\cdot\vec{\sigma}} = 0}$ with $\ds{\left.\expo{\mu\vec{b}\cdot\vec{\sigma}}\right\vert_{\ \mu\ =\ 0} = \sigma_{0}}$ and $\ds{\left.\partiald{\expo{\mu\vec{b}\cdot\vec{\sigma}}}{\mu} \right\vert_{\ \mu\ =\ 0} = \vec{b}\cdot\vec{\sigma}}$ such that $\ds{\pars{~\mbox{note that}\ \vec{b}\cdot\vec{b} = \pars{5\pi \over 8}^{2}~}}$ \begin{align} \expo{\mu\vec{b}\cdot\vec{\sigma}} & = \cosh\pars{{5\pi \over 8}\,\mu}\sigma_{0} + {8 \over 5\pi}\,\sinh\pars{{5\pi \over 8}\,\mu}\vec{b}\cdot\vec{\sigma} \\[8mm] \mbox{and}\ \expo{\mu\,\mathsf{A}} & = \expo{-3\pi\mu/8}\,\cosh\pars{{5\pi \over 8}\,\mu}\sigma_{0} + {8 \over 5\pi}\,\expo{-3\pi\mu/8}\, \sinh\pars{{5\pi \over 8}\,\mu}\vec{b}\cdot\vec{\sigma} \\[4mm] & = \half\,\exp\pars{{\pi \over 4}\mu} \bracks{\sigma_{0} + {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}} + \half\,\exp\pars{-\pi\mu} \bracks{\sigma_{0} - {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}} \end{align}

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\begin{align} A^{n} & = n!\bracks{\mu^{n}}\expo{\mu\,\mathsf{A}} = \half\pars{\pi \over 4}^{n}\bracks{% \sigma_{0} + {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}} + \half\pars{-\pi}^{n}\bracks{% \sigma_{0} - {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}} \end{align}

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\begin{align} \color{#f00}{\sin\pars{A}} & = \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}!}\, A^{2n + 1} = \half\sin\pars{\pi \over 4}\bracks{% \sigma_{0} + {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}} + \half\sin\pars{-\pi}\bracks{% \sigma_{0} - {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}} \\[4mm] & = \half\,{\root{2} \over 2}\bracks{\sigma_{0} + {8 \over 5\pi}\,\pars{A + {3\pi \over 8}\,\sigma_{0}}} = {\root{2} \over 4}\ \underbrace{% \bracks{{8 \over 5}\,\sigma_{0} + {8 \over 5\pi}\,A}} _{\ds{{2 \over 5} \pars{\begin{array}{cc}\ds{1} & \ds{2}\\ \ds{2} & \ds{4}\end{array}}}} \\[3mm] & = \color{#f00}{{1 \over 5} \pars{\begin{array}{cc}\ds{1 \over \root{2}} & \ds{\root{2}} \\[2mm] \ds{\root{2}} & \ds{2\root{2}} \end{array}}} \end{align}