19.4: Finding the Eigenvectors
- Page ID
- 29522
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Now let’s look at the eigenvectors, we’ll start with those of \(P\).Let’s call the eigenvalue \(\lambda\)
Then for an eigenstate of the shift operator, the shifted vector must be just a multiple of the original vector:
\begin{equation}\left(\begin{array}{cccc}0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1 \\1 & 0 & 0 & 0\end{array}\right)\left(\begin{array}{c}A_{1} \\A_{2} \\A_{3} \\A_{4}\end{array}\right)=\left(\begin{array}{c}A_{2} \\A_{3} \\A_{4} \\A_{1}\end{array}\right)=\lambda\left(\begin{array}{c}A_{1} \\A_{2} \\A_{3} \\
A_{4}\end{array}\right)\end{equation}
Reading off the element by element equivalence of the two vectors,
\begin{equation}
A_{2}=\lambda A_{1}, \quad A_{3}=\lambda A_{2}, \quad A_{4}=\lambda A_{3}, \quad A_{1}=\lambda A_{4}
\end{equation}
The first three equalities tell us the eigenvector has the form \(\left(1, \lambda, \lambda^{2}, \lambda^{3}\right)^{T}\), the last tells us that \(\lambda^{4}=1\).
From our earlier discussion of circulant matrices, writing the smallest phase nontrivial \(N^{\text {th }}\) root of unity as \(\omega=e^{2 \pi i / N}\), the roots of the equation \(\lambda^{N}=1\) are just this basic root raised to N different powers: the roots are \(1, \omega, \omega^{2}, \omega^{3}, \ldots, \omega^{N-1}\)
This establishes that the eigenvectors of \(P\) have the form
\begin{equation}
\left(1, \omega^{j},\left(\omega^{j}\right)^{2},\left(\omega^{j}\right)^{3}, \ldots,\left(\omega^{j}\right)^{N-1}\right)^{T}
\end{equation}
where \(j=0,1,2,3, \ldots, N-1\) with corresponding eigenvalue the basic root raised to the \(j^{\text {th }} \text { power, } \omega^{j}=e^{2 \pi i j / N}\). Try it out for 3×3: the eigenvalues are given by
\begin{equation}
\left|\begin{array}{ccc}
-\lambda & 1 & 0 \\
0 & -\lambda & 1 \\
1 & 0 & -\lambda
\end{array}\right|=0, \quad \lambda^{3}=1, \quad \lambda=1, \omega, \omega^{2} ; \quad \omega^{3}=1
\end{equation}
The corresponding eigenvectors are found to be
\begin{equation}
\left(\begin{array}{l}
1 \\
1 \\
1
\end{array}\right), \quad\left(\begin{array}{c}
1 \\
\omega \\
\omega^{2}
\end{array}\right), \quad\left(\begin{array}{c}
1 \\
\omega^{2} \\
\omega
\end{array}\right)
\end{equation}
For the \(N×N\) case, there are \(N\) different, linearly independent, vectors of this form, so this is a complete set of eigenvectors of \(P\).
They are also, of course, eigenvectors of \(\begin{equation}
P^{2}, P^{3} \text { all } N-1 \text { powers of } P
\end{equation}\) and therefore of all the circulant matrices! This means that all \(N×N\)circulant matrices commute.