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        riddles >> putnam exam (pure math) >> Have you seen this kind of matrix?
        
(Message started by: cuckoo on Mar 27^th, 2009, 8:28am)

Title: Have you seen this kind of matrix?
Post by cuckoo on Mar 27^th, 2009, 8:28am

The matrix M is built from two vectors: a[1, m] and b[1, n]. M(i,j)=a(i)+b(j).
Do you know any property about this kind of matrices? 3x!
:D

Title: Re: Have you seen this kind of matrix?
Post by towr on Mar 27^th, 2009, 11:29am

The matrix is the addition of two rank 1 matrices and so has rank 1 or 0. And I think that means it at most has one non-zero eigenvalue.

Title: Re: Have you seen this kind of matrix?
Post by Obob on Mar 27^th, 2009, 2:10pm

The rank can be 2: the vectors [0,1] & [0,1] give the matrix [0,1],[1,2], which is invertible. But the rank can't be any bigger than 2.

Title: Re: Have you seen this kind of matrix?
Post by cuckoo on Mar 27^th, 2009, 8:24pm

you are right.
Thank you all, guys! :D

on 03/27/09 at 14:10:44, Obob wrote:

The rank can be 2: the vectors [0,1] & [0,1] give the matrix [0,1],[1,2], which is invertible. But the rank can't be any bigger than 2.

Title: Re: Have you seen this kind of matrix?
Post by cuckoo on Mar 27^th, 2009, 8:31pm

conversely, if the rank of a matrix is less than or equal to 2, can it be represented in the form [a_i+b_j] ?

Title: Re: Have you seen this kind of matrix?
Post by Obob on Mar 27^th, 2009, 10:18pm

No; for instance, viewing it as a map R^n->R^m, the vector (1,1,...,1) is always in the image of both it and its transpose. This is not a property of all matrices of rank at most 2.

Title: Re: Have you seen this kind of matrix?
Post by Eigenray on Mar 27^th, 2009, 11:33pm

If it's a square matrix, to compute the characteristic polynomial we need only find the coefficient of t^n-2. Thus
det( M - t I ) = (-t)^n-2 [ t² - S(a+b) t + S(a)S(b) - n <a,b> ],
where S(x) = http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/sum.gif x_i.