# Calculate the characteristic polynomial, eigenvalue and eigenvector for this $n\times n$ matrix over $\mathbb{R}$

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*}

• 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. Feb 2, 2023 at 6:42
• @UmeshShankar could you give more details, please Feb 2, 2023 at 6:55
• 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). Feb 2, 2023 at 7:18
• It is a circulant matrix, whose eigenvalues are well known. Feb 2, 2023 at 7:53

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.