19.4: Finding the Eigenvectors

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.