Do you know these matrices?

I have come across a class of matrices with some interesting properties. I feel that they must be known but I have not been able to find anything written about them. This is probably just because I do not know the right place to look. I will describe these matrices here and I hope that somebody will be able to point out a source where I can find more information about them. Consider an $n\times n$ matrix $A$ with elements $a_{ij}$ having the following properties. The elements with $i=j$ (call them $b_i$ ) are negative. The elements with $j=i+1\ {\rm mod}\ n$ (call them $c_i$ ) are positive. All other elements are zero. The determinant of a matrix of this type is $\prod_i b_i+(-1)^{n+1}\prod_i c_i$ . Notice that the two terms in this sum always have opposite signs. A property of these matrices which I found surprising is that $B=(-1)^{n+1}(\det A)A^{-1}$ is a positive matrix, i.e. all its entries $b_{ij}$ are positive. In proving this it is useful to note that the definition of the class is invariant under cyclic permutation of the indices. Therefore it is enough to show that the entries in the first row of $B$ are non-zero. Removing the first row and the first column from $A$ leaves a matrix belonging to the class originally considered. Removing the first row and a column other than the first from $A$ leaves a matrix where $a_{n1}$ is alone in its column. Thus the determinant can be expanded about that element. The result is that we are left to compute the determinant of an $(n-2)\times (n-2)$ matrix which is block diagonal with the first diagonal block belonging to the class originally considered and the second diagonal block being the transpose of a matrix of that class. With these remarks it is then easy to compute the determinant of the $(n-1)\times (n-1)$ matrix resulting in each of these cases. In more detail $b_{11}=(-1)^{n+1}b_2b_3\ldots b_n$ and $b_{1j}=(-1)^{n-j}b_2b_3\ldots b_{j-1}c_j\ldots c_n$ for $j>1$ .

Knowing the positivity of $(-1)^{n+1}(\det A)A^{-1}$ means that it is possible to apply the Perron-Frobenius theorem to this matrix. In the case that $\det A$ has the same sign as $(-1)^{n+1}$ it follows that $A^{-1}$ has an eigenvector all of whose entries are positive. The corresponding eigenvalue is positive and larger in magnitude than any other eigenvalue of $A^{-1}$ . This vector is also an eigenvalue of $A$ with a positive eigenvalue. Looking at the characteristic polynomial it is easy to see that if $(-1)^n(b_1b_2\ldots b_n+(-1)^{n+1}c_1c_2\ldots c_n)<0$ the matrix $A$ has exactly one positive eigenvalue and that none of its eigenvalues is zero.

This entry was posted on March 9, 2012 at 6:41 am and is filed under Uncategorized. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

6 Responses to “Do you know these matrices?”

Rod Carvalho Says:
March 9, 2012 at 8:16 am | Reply
Matrix $A$ is Toeplitz. It could perhaps be viewed as the Laplacian matrix of a weighted directed graph (this graph would be a 1-dimensional “chain” where the “last” vertex has an outward edge incident on the “first” vertex).
Rod Carvalho Says:
March 9, 2012 at 8:18 am | Reply
Sorry. $A$ is certainly not Toeplitz…
hydrobates Says:
March 9, 2012 at 9:04 am | Reply
It is a pity that the Toeplitz suggestion does not work in general. There is nevertheless a non-trivial intersection between the matrices I was interested in and the Toeplitz ones and so there might be something to be learned in that direction.
Luis Guzman Says:
March 14, 2012 at 2:19 am | Reply
Reblogged this on Guzman's Mathematics Weblog and commented:
Do you know these matrices described by Alan Rendall? If so, please point out a source where he may find more information about them. I am interested in knowing too!
anon Says:
March 18, 2012 at 5:12 am | Reply
You might consider posting this on math overflow

http://mathoverflow.net/
- hydrobates Says:
  March 18, 2012 at 8:59 am | Reply
  Thanks for the suggestion. I had never put anything on math overflow before and I have taken this opportunity to try it.