Finding the eigenvalues of an $n \times n$ matrix

Question

Compute the eigenvalues of the $n\times n$ matrix M defined by $$(m_{ij}) = \begin{cases} 1 - 1/n & \text{ if } i = j \\ -1/n & \text{ otherwise.} \end{cases} $$

Using Wolfram Alpha, I found a pattern: for an $n \times n$ matrix, the eigenvalues are $0$ (with multiplicity $1$) and $1$ (with multiplicity $ n - 1)$. But how can I prove this fact? I didn't get anywhere with induction.

For example, in the $n = 2$ case, we have that $\text{trace M } = 1$, and $\text{det} M = 0$. So the product of the eigenvalues is $0$ (which means one eigenvalue is $0$) and the sum is $1$, so the other one must be $1$. This doesn't really work for higher orders though.

José Carlos Santos · Accepted Answer · 2018-09-03T17:58:48.860

2

Since $M-\operatorname{Id}$ has rank $1$ (all columns are equal and different from $0$), $1$ is an eigenvalue with multiplicity $n-1$. And since $(1,1,\ldots,1)$ is an eigenvector with eigenvalue $0$, $0$ is the remaining eigenvalue.

edited Sep 03 '18 at 17:58

answered Sep 03 '18 at 17:56

José Carlos Santos

427,504

1

But for $n > 2,$ $1$ has multiplicity $n - 1$. How can I show this? – Sep 03 '18 at 17:58
@stackofhay42 It was a typo. I meant to write $n-1$. – José Carlos Santos Sep 03 '18 at 17:59
Does $d$ represent an eigenvalue here? – Sep 03 '18 at 18:00
@stackofhay42 There's no $d$ in my anser. There's $\operatorname{Id}$, which is the identity matrix. – José Carlos Santos Sep 03 '18 at 18:01
I understand now. Thanks – Sep 03 '18 at 18:02
@stackofhay42 I'm glad I could help. – José Carlos Santos Sep 03 '18 at 18:14
actually, I don't get it. did you get M - Id from the equation det(M - lambda Id) = 0? and how will rank tell us about multiplicity? – Sep 03 '18 at 18:31
@stackofhay42 Yes, that's where $M-\operatorname{Id}$ comes from. And the rank-nullity theorem tells us that$$\operatorname{rank}(M-\operatorname{Id})=1\iff\dim\ker(M-\operatorname{Id})=n-1.$$ – José Carlos Santos Sep 03 '18 at 19:36

score 0 · Answer 2 · answered Sep 03 '18 at 18:08

The matrix is actually $$A = I - \frac{1}{n} \pmb{1}\pmb{1}^T$$ where $\pmb{1}$ is a column vector of size $n$ of all ones. Notice that $$A\pmb{1} = \pmb{1}- \frac{1}{n} \pmb{1}\pmb{1}^T \pmb{1} = \pmb{1}- \pmb{1} = 0$$ So $ \pmb{1}$ is an eigenvector of eigenvalue zero. Now, the other $n-1$ eigenvalues/vectors come from the following observation. Let $(\lambda,x)$ be eigenvector/value pairs, so $$Ax = x - \frac{\sum x_i}{n} \pmb{1} = x - \bar{x} \pmb{1}$$ where $\bar{x}$ is the sample mean of the entries of $x$. The above equation suggests that $(1, x)$ is an eigenvector/value pair if $\bar{x} = 0$. You could build an $n-1$ linearly independent span of vectors satisfying $\sum x_i = 0$. So \begin{align} \lambda_1 &= 1 \\ & \vdots \\ \lambda_{n-1} &= 1 \\ \lambda_{n} &= 0 \end{align} For $n = 1$ (base case), the only eigenvalue is $0$.

Finding the eigenvalues of an $n \times n$ matrix

2 Answers2