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I am trying to solve the following problem:

Consider the $n\times n$ matrix over $\mathbb{R}$ :

\begin{equation*} \left( \begin{array}{cccccc} 1 & -1 & 0 & \cdots & 0 & 0 \newline 0 & 1 & -1 & \cdots & 0 & 0 \newline 0 & 0 & 1 & \cdots & 0 & 0 \newline \cdot & & & & & \newline \cdot & & & & & \newline 0 & 0 & 0 & \cdots & 1 & -1 \newline -1 & 0 & 0 & \cdots & 0 & 1% \end{array}% \right). \end{equation*}

(a) Calculate its characteristic polynomial.

(b) Find an eigenvalue of this matrix and a corresponding eigenvector.

(c) What is the multiplicity of your eigenvalue?

(d) Is this matrix invertible? Why?.

My try for (a) is as follows, but I think it is not right! Also any help for b,c, and d is appreciated.

The characteristic polynomial of a matrix $A$ is given by $\det(A - \lambda I)$, where $I$ is the identity matrix and $\lambda$ is an eigenvalue of $A$. For the given matrix, the characteristic polynomial is

\begin{align*} \det\left(\left( \begin{array}{cccccc} 1 & -1 & 0 & \cdots & 0 & 0 \newline 0 & 1 & -1 & \cdots & 0 & 0 \newline 0 & 0 & 1 & \cdots & 0 & 0 \newline \cdot & & & & & \newline \cdot & & & & & \newline 0 & 0 & 0 & \cdots & 1 & -1 \newline -1 & 0 & 0 & \cdots & 0 & 1% \end{array}% \right) - \lambda I\right) \\ = \det\left( \begin{array}{cccccc} 1-\lambda & -1 & 0 & \cdots & 0 & 0 \newline 0 & 1-\lambda & -1 & \cdots & 0 & 0 \newline 0 & 0 & 1-\lambda & \cdots & 0 & 0 \newline \cdot & & & & & \newline \cdot & & & & & \newline 0 & 0 & 0 & \cdots & 1-\lambda & -1 \newline -1 & 0 & 0 & \cdots & 0 & 1-\lambda% \end{array}% \right) \end{align*} \begin{align*} = (1-\lambda)^{n-1} \det\left( \begin{array}{ccccc} 1-\lambda & -1 \newline -1 & 1-\lambda% \end{array}% \right) \newline = (1-\lambda)^{n-1} [(1-\lambda)^2 + 1]. \end{align*}

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    $\begingroup$ The matrix you have can be written as $A=I-P$ where $P$ is a permutation matrix. It is also quite a nice permutation, and you can find it's characteristic polynomial quite easily. $\endgroup$ Feb 2, 2023 at 6:42
  • $\begingroup$ @UmeshShankar could you give more details, please $\endgroup$
    – Math Lover
    Feb 2, 2023 at 6:55
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    $\begingroup$ Note that $P^n =I$ so $(I-A)^n = I$. For (b) consider the vector of ones. Part (c) follows from (a) and (d) follows from (b). $\endgroup$
    – copper.hat
    Feb 2, 2023 at 7:18
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    $\begingroup$ It is a circulant matrix, whose eigenvalues are well known. $\endgroup$
    – Jean Marie
    Feb 2, 2023 at 7:53

1 Answer 1

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Say $n\gt1.$ The characteristic polynomial is $$p_A(\lambda) =(1-\lambda) ^n-1.$$

So $\lambda =0$ is an eigenvalue, of multiplicity $1$. An eigenvector is $\begin {pmatrix}1\\1\\\vdots \\1\end{pmatrix}.$

The determinant is $0$, since $0$ is an eigenvalue.

And $A$ is not invertible.

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