1.6: Matrix Representation of Complex Numbers
- Page ID
- 96141
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)In our studies of complex numbers, we noted that multiplication of a complex number by \(e^{iθ}\) rotates that complex number an angle \(θ\) in the complex plane. This leads to the idea that we might be able to represent complex numbers as matrices with \(e^{iθ}\) as the rotation matrix.
Accordingly, we begin by representing \(e^{iθ}\) as the rotation matrix, that is,
\[\begin{aligned}e^{i\theta}&=\left(\begin{array}{rr}\cos\theta&-\sin\theta \\ \sin\theta&\cos\theta\end{array}\right) \\ &=\cos\theta\left(\begin{array}{cc}1&0\\0&1\end{array}\right)+\sin\theta\left(\begin{array}{rr}0&-1\\1&0\end{array}\right).\end{aligned} \nonumber \]
Since \(e^{iθ} = \cos θ + i \sin θ\), we are led to the matrix representations of the unit numbers as
\[1=\left(\begin{array}{cc}1&0\\0&1\end{array}\right),\quad i=\left(\begin{array}{rr}0&-1\\1&0\end{array}\right).\nonumber \]
A general complex number \(z = x + iy\) is then represented as
\[z=\left(\begin{array}{rr}x&-y\\y&x\end{array}\right).\nonumber \]
The complex conjugate operation, where \(i → −i\), is seen to be just the matrix transpose.
Show that \(i^2=-1\) in the matrix representation.
Solution
We have
\[i^2=\left(\begin{array}{rr}0&-1\\1&0\end{array}\right)\left(\begin{array}{rr}0&-1\\1&0\end{array}\right)=\left(\begin{array}{rr}-1&0\\0&-1\end{array}\right)=-\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=-1.\nonumber \]
Show that \(z\overline{z}=x^2+y^2\) in the matrix representation.
Solution
We have
\[z\overline{z}=\left(\begin{array}{rr}x&-y\\y&x\end{array}\right)\left(\begin{array}{rr}x&y\\-y&x\end{array}\right)=\left(\begin{array}{cc}x^2+y^2&0\\0&x^2+y^2\end{array}\right)=(x^2+y^2)\left(\begin{array}{cc}1&0\\0&1\end{array}\right)=(x^2+y^2).\nonumber \]
We can now see that there is a one-to-one correspondence between the set of complex numbers and the set of all two-by-two matrices with equal diagonal elements and opposite signed off-diagonal elements. If you do not like the idea of \(\sqrt{-1}\), then just imagine the arithmetic of these two-by-two matrices!