Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Please help me calculate eigenvalues and eigenvectors for a square matrix, with $n$ rows and $n$ columns, and $-1$ in each cell:$$\begin{bmatrix} -1 & \,\,\cdots & -1\\\,\,\vdots \strut& \,\,\ddots & \,\,\vdots \strut\\ -1 &\,\,\cdots & -1\end{bmatrix}$$

I guess that this is a $0$ and $n$, vectors $(1,1,\ldots,1)$ and $(1,-1,0,\ldots)$, $(1,0,-1,0,\ldots),\ldots$

How to prove it?

share|cite|improve this question
To be clear, you mean a matrix of the form $$\begin{bmatrix} -1 & \cdots & -1\\\,\vdots \strut& \,\ddots & \,\vdots \strut\\ -1 &\cdots & -1\end{bmatrix}\qquad ?$$ – Zev Chonoles Feb 28 '13 at 20:00
Yes, that's correct matrix – Steve Feb 28 '13 at 20:03
up vote 2 down vote accepted

For $i=1\ldots n$ let $e_i$ be the $i$-th vector of the canonical basis of $\mathbb R^n$, i.e. $$ (\vec e_i)_j = \begin{cases} 1 & \text{if }i=j \\ 0 & \text{if }i\neq j\end{cases} $$ If $M_{ij}=-1$ $\forall i,j$ then, as you say, $\vec v_i=\vec e_1-\vec e_i$ ($i=2\ldots n$) are all eigenvectors of $M$ with eigenvalue $0$. In fact for all $j=1\ldots n$ $$ (M\vec v_i)_j = \sum_{\ell=1}^n M_{j\ell}(\vec v_i)_\ell = (-1)\cdot (\vec v_i)_1 + (-1)\cdot(\vec v_i)_i = -1 -(-1) = 0 $$ Therefore $M\vec v_i = \vec 0 = 0\cdot\vec v_i$.

A similar computation shows that $\vec u=(1\ldots 1)$ is an eigenvector of $M$ with eigenvalue $1$.

$\vec u,\vec v_2\ldots \vec v_n$ are $n$ linearly independent eigenvectors, i.e. they form a basis of $\mathbb R^n$ so no more linearly independent eigenvectors exist.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.