# Does this matrix shape have a name?

I'm using a lot of matrices that look like this: $$A_3 = \begin{bmatrix} a & b & b\\ b & a & b\\ b & b & a \end{bmatrix}$$

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!

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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 –  Gilead Mar 31 '13 at 19:28
Why not even dimension? I would say $aI+b(J-I)$ or $(a-b)I+bJ$ –  Aaron Meyerowitz Mar 31 '13 at 19:30
I'm not interested in even dimension right now. The name should hold for any dimension - so my question had superfluous information. –  Nathaniel Schwartz Mar 31 '13 at 19:36
@Gilead, that link is the kind of thing I am looking for; but nothing there matches my matrix, as far as I can see. –  Nathaniel Schwartz Mar 31 '13 at 19:45
By the way, for this $n \times n$ matrix to be non-singular you also need $a + (n-1)b \ne 0$. –  Robert Israel Mar 31 '13 at 20:41

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.)

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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 $n=2$. –  Robert Israel Mar 31 '13 at 20:51
Thanks. I will try to understand how the Bose Mesner algebra relates to these matrices. As yet, I don't know. –  Nathaniel Schwartz Mar 31 '13 at 21:10
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! –  Suvrit Mar 31 '13 at 21:29
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. –  Chris Godsil Mar 31 '13 at 23:59
@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 $K_n$ –  Suvrit Apr 1 '13 at 1:03

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.

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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. –  Igor Khavkine Apr 1 '13 at 19:49

Circulant Matrices? What is your motivation?

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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. –  Nathaniel Schwartz Mar 31 '13 at 21:21

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

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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.

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