mathematics | VictorX

What’s eigenvalues of a $n\times n$ matrix $A=\begin{pmatrix} 1 & c & c & \dots & c\\ c & 1 & c & \dots & c\\ c & \vdots &\vdots && \vdots \\ \vdots &\vdots & & 1&c\\ c&c&c\dots&c&1\end{pmatrix}$ ?

Solution 1: Notice that $A=c J+(1-c)I$ , where $I,J$ are identity and ones matrix respectively. Assume $v$ is an eigen-vector associated with an eigen-value $\lambda$ , then

$\begin{aligned} &&cJv+(1-c)Iv=\lambda v\\ &\implies& c|v|\overset{\rightarrow}{1}=(\lambda+c-1) v. \end{aligned}$

Therefore, $v=\overset{\rightarrow}{1}$ , when $c|v|=(\lambda+c-1)$ , i.e., $\lambda_1=1+(n-1)c$ .

Otherwise, $|v|=0$ when $\lambda_{2,\dots,n}=1-c$ , this is a $n-1$ dimension subspace $\text{span}\{\begin{pmatrix} 1\\,-1\\, 0\\, \vdots,\\0\end{pmatrix}, \begin{pmatrix} 1\\,0\\, -1\\, \vdots,\\0\end{pmatrix}, \dots, \begin{pmatrix} 1\\,0\\, 0\\, \vdots,\\-1\end{pmatrix}\}$ .

Solution 2: Matrix Determinant lemma (wikipedia: https://en.wikipedia.org/wiki/Matrix_determinant_lemma#Proof)

Statement: $\text{det} (A+uv^T)=(1+v^TA^{-1}u)\text{det}(A)$ , where $A$ is an invertible matrix, $u,v\in \mathbb{R}^{n,1}$ .

Let’s denote $A=(1-c-\lambda)I_{n,n}$ , and $u,v=\sqrt c \overset{\rightarrow}{1}_{n,1}$ , we can solve the eigenvalue from the corresponding polynomial

$\begin{aligned} (1+\frac{c}{1-c-\lambda}\cdot n)(1-c-\lambda)^n=(1-\lambda+(n-1)c)(1-c-\lambda)^{n-1} \end{aligned}$ .

Therefore, the eigenvalues $\lambda_1=1+(n-1)c, \lambda_{2,\dots,n}=1-c$ .

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