2
$\begingroup$

I'm using a lot of matrices that look like the following

$$ A_3 = \begin{bmatrix} a & b & b\\ b & a & b\\ b & b & a \end{bmatrix} $$

i.e., the diagonal entries are all the same, and all off-diagonal entries are the same. They need to be non-singular, so $a \ne b$, for one thing.

If you know of a common name for this shape of matrix, I would very much appreciate your help, and references would be most appreciated!

Improve this question
$\endgroup$
7
  • $\begingroup$ It looks like a special-case of a symmetric matrix, but I'm not sure if there's a specific name for it. You might want to go through the list here: en.wikipedia.org/wiki/List_of_matrices $\endgroup$
    – Gilead
    Commented Mar 31, 2013 at 19:28
  • 2
    $\begingroup$ Why not even dimension? I would say $aI+b(J-I)$ or $(a-b)I+bJ$ $\endgroup$
    – Aaron Meyerowitz
    Commented Mar 31, 2013 at 19:30
  • $\begingroup$ I'm not interested in even dimension right now. The name should hold for any dimension - so my question had superfluous information. $\endgroup$
    – Nathaniel Schwartz
    Commented Mar 31, 2013 at 19:36
  • $\begingroup$ @Gilead, that link is the kind of thing I am looking for; but nothing there matches my matrix, as far as I can see. $\endgroup$
    – Nathaniel Schwartz
    Commented Mar 31, 2013 at 19:45
  • $\begingroup$ By the way, for this $n \times n$ matrix to be non-singular you also need $a + (n-1)b \ne 0$. $\endgroup$
    – Robert Israel
    Commented Mar 31, 2013 at 20:41

5 Answers 5

Reset to default
12
$\begingroup$

As far as I am aware, these matrices are called Bose Mesner matrices--For reasons that I now do not remember. However, many years ago, I wrote a short summary that is still available at this link

(In particular, they form a nice algebra, their characteristic polynomial has a closed form, etc. etc.)

Improve this answer
$\endgroup$
7
  • 2
    $\begingroup$ Looking at en.wikipedia.org/wiki/Bose%E2%80%93Mesner_algebra it appears that Bose-Mesner matrices are somewhat more general: I think you have the case $n=2$. $\endgroup$
    – Robert Israel
    Commented Mar 31, 2013 at 20:51
  • $\begingroup$ Thanks. I will try to understand how the Bose Mesner algebra relates to these matrices. As yet, I don't know. $\endgroup$
    – Nathaniel Schwartz
    Commented Mar 31, 2013 at 21:10
  • $\begingroup$ Thanks Robert---I was waiting for a comment from you, because somehow I get the feeling that 10yrs ago I heard of this name from you on sci.math or something :-) --- and of course, the general class of Bose-Mesner matrices or algebra is much richer than this special example, but this at least starts us out with some name! $\endgroup$
    – Suvrit
    Commented Mar 31, 2013 at 21:29
  • $\begingroup$ In the context of association schemes, these matrices are not called Bose-Mesner matrices. The matrices given do belong to the Bose-Mesner algebra of the complete graph, but I have never seen any matrix in any Bose-Mesner algebra referred to as a Bose-Mesner matrix. $\endgroup$
    – Chris Godsil
    Commented Mar 31, 2013 at 23:59
  • $\begingroup$ @Chris: I think I then just extrapolated from the BM algebra to call these matrices BM matrices! But given the connection, perhaps it is acceptable to call them: BM matrices of type $K_n$ $\endgroup$
    – Suvrit
    Commented Apr 1, 2013 at 1:03
3
$\begingroup$

Notice that a matrix with constant entries $b$ can be viewed as $b$ times the projector onto the vector $v=(1,\ldots,1)$. Thus your matrix is

\begin{align} A_n=(a-b)\mathbb{I}_n+b \,v v^\top. \end{align}

Thus its eigenvalues are $a$, and $a-b$, and eigenvectors are $v$ and all vectors orthogonal to $v$.

Incidentally, $A_n$ commutes with all permutation matrices of the $n$-element permutation group in the $n$ dimensional representation. Thus it is a Casimir of that representation.

Improve this answer
$\endgroup$
1
  • 2
    $\begingroup$ In numerical analysis, the addition of a $v v^T$ term is often called a rank-1 update. Thus, $A_n$ is a rank-1 update of a multiple of the identity. $\endgroup$
    – Igor Khavkine
    Commented Apr 1, 2013 at 19:49
0
$\begingroup$

Circulant Matrices? What is your motivation?

Improve this answer
$\endgroup$
1
  • $\begingroup$ I'm studying automorphisms of order 2 on certain algebraic groups. These matrices come up in that context. Circulant matrices are more general, though these certainly fit that shape. $\endgroup$
    – Nathaniel Schwartz
    Commented Mar 31, 2013 at 21:21
0
$\begingroup$

if $a > b$ then the matrix is a special case of a Robinsonian matrix or R-Matrix; I encountered that name when searching for "matrix reordering" and remembered the question on MathOverflow

vis.pku.edu.cn/file/Summer%20School%202009/feket_matrix.pdf

Improve this answer
$\endgroup$
-1
$\begingroup$

I think the best description would be a Circulant Matrix, as the other "unknown (google)" suggests since what you are seeing is the vector (a b .... b) shifting right 1 on each row.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .