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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 ?

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    what is $a_{11}$. how can the trace be 0 – 2017-02-08
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    $a_{1,1}=0$ since it is one of the "Other" entries, right? – 2017-02-08
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    @Upstart , since all the entries in the main diagonal is $0$ , hence trace is $0$. – 2017-02-08
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    @Laray yes it is $0$. – 2017-02-08
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    now it is clear it was not clear before the edit – 2017-02-08

3 Answers 3

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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.

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    can you give some reference this statement - "The set of all the matrices which are just the permutaion of the identity matrix of order $n∗n$ is isomorphic to the group $S_{n}$. " – 2017-02-08
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    any reference for the above statement ? – 2017-02-19
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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.

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    You are right on that, I will change my answer. Thank you! – 2017-02-08
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    @Laray if you see the question carefully then here $n$ is an odd integer $\geq 7$ and we cannot take $n = 4$ here!. – 2017-02-08
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    Sorry, I read over that. It does not change the fact about the characteric polynomial or the roots of unity. This will, on the other hand, make it impossible to have $-1$ as eigenvalue. – 2017-02-08
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    It's the companion matrix of $x^n-1$ – 2017-02-08
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    I hate transposing and weird indexing... You are right, I will correct that in my answer! – 2017-02-08
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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?