Take the 2-minute tour ×
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.

Let $A$ be a $n\times n$ matrix above the reals, with ones in the diagonal, and all of the other elements are equal to $r$ with $0<r<1$.

How can I prove that the eigenvalues of $A$ are $1+(n-1)r$ and $1-r$, with multiplicity $n-1$?

share|improve this question
1  
Any ideas of your own...? –  user86418 Feb 24 at 22:36

3 Answers 3

The matrix $A$ can be expressed as $$ A=ruu^T+(1-r)I $$ where $u=(1,1,\ldots,1)\in\mathbb R^{n\times 1}$, and $I$ the identity matrix in $\mathbb R^{n\times n}$.

So if $Av=\lambda v$, with $v\ne 0$, then $$ \lambda v=Av=r\langle u,v\rangle u+(1-r)v $$ where $\langle\cdot,\cdot\rangle$ is the standard inner product in $\mathbb R^n$. The above implies that $$ (\lambda-1+r)v=r\langle u,v\rangle u, $$ which in turn implies that,

either

  • $\langle u,v\rangle =0$, which happens for $n-1$ linearly independent vectors (as many the the dimension of the perpendicular hyperspace to $u$), and in this case $$ \lambda=1-r, $$ and hence the eigenvlaue $\lambda=1-r$ has multiplicity $n-1$,

or

  • $\langle u,v\rangle \ne 0$, which means that $v$ is a multiple of $u$, and the corresponding eigenvalue is obtained for $v=u$: $$ (\lambda-1+r)u=r\langle u,u\rangle u, $$ i.e., $$ \lambda=r\langle u,u\rangle +1-r=r(n-1)+1. $$
share|improve this answer
    
Really great proof! –  Léo Feb 24 at 22:41
3  
You can simplify your argument a lot if you simply write $$A=r\,uu^T+(1-r)I=nr\,\frac{uu^T}n+(1-r)\frac{uu^T}n+(1-r)(I-\frac{uu^T}n)\\ = (nr+1-r)\,\frac{uu^T}n+(1-r)\,(I-\frac{uu^T}n).$$ Then you read directly that the eigenvalues are $1-r$ with multiplicity $n-1$ and $nr+1-r=1+n(r-1)$ with multiplicity $1$. –  Martin Argerami Feb 24 at 22:59

The determinant is invariant if you add to any row (column) a linear combination of other rows (columns). Then you can first subtract to every row the row right below it and you get \begin{vmatrix} 1 -\lambda& r & r & \cdots & r\\ r & 1 -\lambda & r & \cdots & r\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ r & r & r & \cdots & 1-\lambda \end{vmatrix} equal to \begin{vmatrix} 1-r-\lambda & r-1+\lambda & 0 & \cdots & 0\\ 0 & 1-r-\lambda & r-1+\lambda & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ r & r & r & \cdots & 1-\lambda \end{vmatrix} and now starting on the second column add to each column the one on it right like \begin{vmatrix} 1-r-\lambda & 0 & 0 & \cdots & 0\\ 0 & 1-r-\lambda & r-1+\lambda & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ r & 2r & r & \cdots & 1-\lambda \end{vmatrix} and when you do it for all of them you have \begin{vmatrix} 1-r -\lambda& 0 & 0 & \cdots & 0\\ 0 & 1-r -\lambda& 0 & \cdots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ r & 2r & 3r & \cdots & 1 + (n-1)r-\lambda \end{vmatrix} finally using the cofactor formula for the determinant you get the polynomial $$ (1-r-\lambda)^{n-1}(1+(n-1)r-\lambda)=0 $$ which has the roots you wanted.

share|improve this answer

It's clear that $1 - r $is an eigenvalue whose corresponding eigenspace is $n - 1$ dimensional (it's the null space of a matrix consisting only of $1$s).

On the other hand by inspection the column vector consisting only of 1s is an eigenvector with eigenvalue $1 + (n-1) r $and the multiplicity must be $1$ since we already have $n - 1$.

share|improve this answer

Your Answer

 
discard

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.