How to Find Eigenvalues of a Specific Matrix.

Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra

Problem 23

Find all eigenvalues of the following $n \times n$ matrix.

\[
A=\begin{bmatrix}
0 & 0 & \cdots & 0 &1 \\
1 & 0 & \cdots & 0 & 0\\
0 & 1 & \cdots & 0 &0\\
\vdots & \vdots & \ddots & \ddots & \vdots \\
0 & 0&\cdots & 1& 0 \\
\end{bmatrix}
\]

FavoriteLoadingAdd to solve later

Sponsored Links

Steps.

  1. Use the 1st row cofactor expansion
  2. For each smaller determinant matrix, use cofactor expansion inductively.

Solution.

We calculate the characteristic polynomial $\det(A-\lambda I)$ of $A$.

We use the cofactor expansion corresponding to the first row.
\begin{align*}
&\det(A-\lambda I)=
\begin{vmatrix}
-\lambda & 0 & \cdots & 0 &1 \\
1 & -\lambda & \cdots & 0 & 0\\
0 & 1 & \cdots & 0 &0\\
\vdots & \vdots & \ddots & \ddots & \vdots \\
0 & 0&\cdots & 1& -\lambda \\
\end{vmatrix} \\[6pt] &=-\lambda
\begin{vmatrix}
-\lambda & 0 & \cdots & 0 &0 \\
1 & -\lambda & \cdots & 0 & 0\\
0 & 1 & \cdots & 0 &0\\
\vdots & \vdots & \ddots & \ddots & \vdots \\
0 & 0&\cdots & 1& -\lambda \\
\end{vmatrix}
+(-1)^{n+1}
\begin{vmatrix}
1 & -\lambda & \cdots & 0 &0 \\
0 & 1 & \cdots & 0 & 0\\
\vdots & \vdots & \ddots & \ddots & \vdots \\
0 & 0 & \cdots & 1 & -\lambda \\
0 & 0&\cdots & 0& 1 \\
\end{vmatrix} \\
\end{align*}

Now we calculate the two determinants in the second equality separately.

The easiest way is to note that the determinant of a triangular matrix is the product of its diagonal entries.
Thus the first determinant is $(-\lambda)^{n-1}$ and the second determinant is $1$.

(If you don’t know this fact, then use the first row cofactor expansion inductively to compute the first determinant. For the second one, use the first column cofactor expansion inductively.)

Thus we obtain $\det(A-\lambda I)=(-1)^n\lambda^n+(-1)^{n+1}=(-1)^n(\lambda^n-1)$.
Therefore eigenvalues are $n$-th roots of unity $e^{2\pi i/n}$ for $i=0,1,\dots, n-1$.

Comment.

The original determinant is not in a good shape for induction but once we apply the 1st row cofactor expansion the smaller determinants obtained are better suited for induction.
When I say in the proof “inductively”, I meant that you need to use mathematical induction to prove the claim (more) rigorously.


FavoriteLoadingAdd to solve later

Sponsored Links

More from my site

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

More in Linear Algebra
Problems and Solutions of Eigenvalue, Eigenvector in Linear Algebra
If Every Trace of a Power of a Matrix is Zero, then the Matrix is Nilpotent

Let $A$ be an $n \times n$ matrix such that $\tr(A^n)=0$ for all $n \in \N$. Then prove that $A$...

Close