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)
 

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*}

The post Characteristic Polynomial, Eigenvalues, Diagonalization Problem (Princeton University Exam) first appeared on Problems in Mathematics.

Answer the following two questions with justification.

(a) Does there exist a $2 \times 2$ matrix $A$ with $A^3=O$ but $A^2 \neq O$? Here $O$ denotes the $2 \times 2$ zero matrix.

(b) Does there exist a $3 \times 3$ real matrix $B$ such that $B^2=A$ where
\[A=\begin{bmatrix}
1 & -1 & 0 \\
-1 &2 &-1 \\
0 & -1 & 1
\end{bmatrix}\,\,\,\,?\]

(Princeton University Linear Algebra Exam)

Hint.

1. (a) Consider the eigenvalues and the Jordan canonical form of $A$
2. (b) Diagonalize the matrix $A$

Solution.

(a) Does there exist a $2 \times 2$ matrix $A$ with $A^3=O$ but $A^2 \neq O$?

We claim that there is no such $2 \times 2$ matrix $A$.
Suppose that we have $A^3=O$. This implies that all the eigenvalues of $A$ are zero.
To see this, let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be a corresponding eigenvector. Then applying the defining relation $A\mathbf{x}=\lambda \mathbf{x}$ successively, we have
\[\mathbf{0}=A^3\mathbf{x}=\lambda A^2\mathbf{x}=\lambda^2 A\mathbf{x}=\lambda^3 \mathbf{x}.\] Thus, the eigenvalue $\lambda$ must be zero since $\mathbf{x}$ is a nonzero vector since it is an eigenvector.
Therefore, there exists an invertible matrix $P$ such that the Jordan canonical form of $A$ is $P^{-1}AP=\begin{bmatrix}
0 & a\\
0& 0
\end{bmatrix}$ for some number $a$.
Then $A=P\begin{bmatrix}
0 & a\\
0& 0
\end{bmatrix} P^{-1}$ and we have
\begin{align*}
A^2&=\left( P\begin{bmatrix}
0 & a\\
0& 0
\end{bmatrix} P^{-1}\right) \left(P\begin{bmatrix}
0 & a\\
0& 0
\end{bmatrix} P^{-1}\right)= P\begin{bmatrix}
0 & a\\
0& 0
\end{bmatrix}^2P^{-1} \\[6pt] &=P\begin{bmatrix}
0 & 0\\
0& 0
\end{bmatrix}P^{-1}=O.
\end{align*}
Hence $A^2$ must be zero, and we conclude that there is no $2 \times 2$ matrix $A$ satisfying both $A^3=O$ and $A^2 \neq O$.

(b) Does there exist a $3 \times 3$ real matrix $B$ such that $B^2=A$?

The answer is yes and we will construct such a matrix $B$ as follows.

The characteristic polynomial $p(t)$ of the matrix $A$ is
\[p(t):=\det(A-tI)=-t(t-1)(t-3).\] Thus the eigenvalues are $\lambda=0, 1, 3$.
The corresponding eigenvalues are obtained by solving the system $(A-\lambda I)\mathbf{x}=\mathbf{0}$ and we see that
\[\begin{bmatrix}
1 \\
1 \\
1
\end{bmatrix}, \begin{bmatrix}
1 \\
0 \\
-1
\end{bmatrix}, \begin{bmatrix}
1 \\
-2 \\
1
\end{bmatrix}\] are eigenvectors corresponding to eigenvalues $0,1,3$, respectively.

Thus the matrix
\[P=\begin{bmatrix}
1 & 1 & 1 \\
1 &0 &-2 \\
1 & -1 & 1
\end{bmatrix}\] is invertible and diagonalize the matrix $A$ such that
\[P^{-1}AP=\begin{bmatrix}
0 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 3
\end{bmatrix}.\]

Then if we have the equality $B^2=A$, then we need to have
\[(P^{-1}BP)(P^{-1}BP)=p^{-1}AP=\begin{bmatrix}
0 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 3
\end{bmatrix}.\] From this we see that the matrix
\[B=P\begin{bmatrix}
0 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & \sqrt{3}
\end{bmatrix}P^{-1}\] satisfies the above relation.

In fact, we compute
\begin{align*}
B^2&=\left( P\begin{bmatrix}
0 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & \sqrt{3}
\end{bmatrix}P^{-1} \right) \left(P\begin{bmatrix}
0 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & \sqrt{3}
\end{bmatrix}P^{-1}\right) \\[6pt] &=P\begin{bmatrix}
0 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & \sqrt{3}
\end{bmatrix}^2P^{-1}=P\begin{bmatrix}
0 & 0 & 0 \\
0 &1 &0 \\
0 & 0 & 3
\end{bmatrix}P^{-1}=A.
\end{align*}
Remark that the matrix $B$ is real since $P$ is real. Thus the matrix $B$ satisfies the conditions of the problem.

Remark.

The explicit form of the matrix $P$ was used to assure that the matrix $B$ is real.
So we did not have to compute the matrix $P$ explicitly if we use the fact that the symmetric matrix is diagonalizable by a real orthogonal matrix.

Generalization.

See problem A square root matrix of a symmetric matrix with non-negative eigenvalues for a more general question than part (b).

Furthermore, try the next problem.

For a solution of this problem, see the post
A Positive Definite Matrix Has a Unique Positive Definite Square Root

The post A Square Root Matrix of a Symmetric Matrix first appeared on Problems in Mathematics.