Eigenvalues of matrix $A$ where $A$ is an $n$ by $n$ matrix where $a_{i,i+1} = 1 \forall i=1,2,...,n-1 $,$a_{n,1}=1$,Other remaining entries are $0$.

Question

Let $A$ be a $n$ by $n$ matrix where $n$ is odd and $n \geq 7$ , and $a_{i,i+1} = 1 \forall i=1,2,...,$ and $a_{n,1}=1$ ,additionally other entries are $0$, i.e. $$A=\begin{bmatrix} 0 & 1 & 0 & 0 & \cdots & 0 \\ 0 & 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & 1 \\ 1 & 0 & 0 & 0 & \cdots & 0 \end{bmatrix}$$ Can $A$ have eigen value of $-1$, also can $A$ have an eigenvalue of multiplicity $\geq2$?

I tried that $Trace(A) = 0$ and $Det(A) = 1$ from the structure of matrix $A$ , i concluded that $0 $ cannot be an eigenvalue ,and from taking a particular case of $n=7$ I got eigenvalue of $1$, now how do i proceed ?

Answer 1

The set of all the matrices which are just the permutaion of the identity matrix of order $n*n$ is isomorphic to the group $S_n$. Consider your matrix as an element of $S_n$. The matrix can be regarded as the permutation ($1$ $n$ $n-1$ $n-2$.......$2$). This is a $n-cycle$. hence its order is $n$. That means $x^n=idt$ where $idt$ is the identity permutation.

Hence your matrix has characteristic equation $x^n-1=0$.

If $n$ is odd then $-1$ is not an eigenvalue.If $n$ is even then $-1$ is an eigenvalue. And since these are the $n$th roots of unity they can't be repeated.

Answer 2

You have $A = \begin{pmatrix} 0&1&0\\0 &0&1\\1&0&0\end{pmatrix}$ for $n=3$, right?

This matrix is the transpose of the Companion matrix of a quite simple polynomial $p=x^n -1$

Since the matrix has only a single entry in every row and column, it's orthogonal and it's eigenvalues are the $n$-th roots of unity.

Answer 3

Hint: What you are describing is a special case of a Circulant matrix. A general circulant matrix takes the form $$ \begin{pmatrix} c_0 & c_{n-1} & \dots & c_2 & c_1 \\ c_1 & c_0 & c_{n-1} & & c_2 \\ \vdots & c_1 & c_0 & \ddots & \vdots \\ c_{n-2} & & \ddots & \ddots & c_{n-1} \\ c_{n-1} & c_{n-2} & \dots & c_1 & c_0 \end{pmatrix} $$ and has eigenvalues $$ \lambda_j = c_0 + c_{n-1} \omega_j + c_{n-1} \omega_j^2 + \dots + c_1 \omega_j^{n-1}, \quad j=0, \dots, n-1, $$ where $\omega_j$ are the $n$th roots of unity. What is $c_k, \, k=0, \dots n-1$ in your case? What does this tell you about the eigenvalues of your matrices?