Characteristic Polynomial, Eigenvalues, Diagonalization Problem (Princeton University Exam)

Problem 178

Let
\[\begin{bmatrix}
0 & 0 & 1 \\
1 &0 &0 \\
0 & 1 & 0
\end{bmatrix}.\]

(a) Find the characteristic polynomial and all the eigenvalues (real and complex) of $A$. Is $A$ diagonalizable over the complex numbers?

(b) Calculate $A^{2009}$.

(Princeton University, Linear Algebra Exam)
 
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Solution.

(a) The characteristic polynomial and the eigenvalues

The characteristic polynomial $p(t)$ of the matrix $A$ is the determinant of $A-tI$. We compute $p(t)=\det(A-tI)$ as follows.
\begin{align*}
p(t)&=\det(A-tI)\\
&=\begin{vmatrix}
-t & 0 & 1 \\
1 &-t &0 \\
0 & 1 & -t
\end{vmatrix}\\
&=-t\begin{vmatrix}
-t & 0\\
1& -t
\end{vmatrix}+\begin{vmatrix}
1 & -t\\
0& 1
\end{vmatrix} \text{ by the first row cofactor expansion}\\
=-t^3+1.
\end{align*}

Thus, the characteristic polynomial of the matrix $A$ is
\[p(t)=-t^3+1.\] The eigenvalues of the matrix $A$ is roots of the characteristic polynomial.
Hence solving $-t^3+1=0$, we obtain
\[t=1, \frac{-1\pm\sqrt{3}i}{2}\] and these are all eigenvalues of $A$.

The matrix $A$ is a $3\times 3$ matrix, and hence it has at most three distinct eigenvalues, and
we found three distinct eigenvalues. In general, if an $n\times n$ matrix has $n$ distinct eigenvalues, the matrix is diagonalizable. Thus the matrix $A$ is diagonalizable.

(b) $A^{2009}$

By direct computations, we have
\[A^2=\begin{bmatrix}
0 & 0 & 1 \\
1 &0 &0 \\
0 & 1 & 0
\end{bmatrix}\begin{bmatrix}
0 & 0 & 1 \\
1 &0 &0 \\
0 & 1 & 0
\end{bmatrix}=\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
1 & 0 & 0
\end{bmatrix}\] \[A^3=AA^2=\begin{bmatrix}
0 & 0 & 1 \\
1 &0 &0 \\
0 & 1 & 0
\end{bmatrix}\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
1 & 0 & 0
\end{bmatrix}=I,\] where $I$ is the $3\times 3$ identity matrix.

Therefore noting that $2009=3\cdot 669 +2$, we have
\begin{align*}
A^{2009}&=A^{3\cdot 669+2}=(A^{3})^{669} A^2\\
&=I^{669}A^2=A^2.
\end{align*}
Therefore, we obtain
\[A^{2009}=\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
1 & 0 & 0
\end{bmatrix}.\]

Another solution using diagonalization

Here, I give another solution for (b) using the diagonalization of the matrix $A$.
For this particular matrix $A$, the above solution is easier since the power of $A$ has a simple pattern.

The following computation is lengthy, but I give it for a pedagogical reason.
I just outline the solution and omit the detail computations.
We obtained eigenvectors $\zeta^{k}, k=0,1,2$, where $\zeta$ is a primitive third root of unity in (a).
Eigenvectors are $\begin{bmatrix}
\zeta^{2k} \\
\zeta^k \\
1
\end{bmatrix}$
for each eigenvalue $\zeta^k$.
Then let
\[S=\begin{bmatrix}
1 & \zeta^2 & \zeta \\
1 &\zeta &\zeta^2 \\
1 & 1 & 1
\end{bmatrix}\] be the matrix whose columns are eigenvectors.

It has the inverse
\[S^{-1}=\frac{1}{3}\begin{bmatrix}
1 & 1 & 1 \\
\zeta &\zeta^2 &1 \\
\zeta^2 & \zeta & 1
\end{bmatrix}.\] Then the matrix $S$ diagonalize the matrix $A$ and we obtain
\[S^{-1}AS=\begin{bmatrix}
1 & 0 & 0 \\
0 &\zeta &0 \\
0 & 0 & \zeta^2
\end{bmatrix}.\] Then we compute
\begin{align*}
A^{2009}&=S\begin{bmatrix}
1 & 0 & 0 \\
0 &\zeta^{2009} &0 \\
0 & 0 & \zeta^{2(2009)}
\end{bmatrix} S^{-1}\\
&=S\begin{bmatrix}
1 & 0 & 0 \\
0 &\zeta^2 &0 \\
0 & 0 & \zeta
\end{bmatrix}S^{-1}
=\begin{bmatrix}
0 & 1 & 0 \\
0 &0 &1 \\
1 & 0 & 0
\end{bmatrix}.
\end{align*}

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