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Topic: Have you seen this kind of matrix? (Read 1286 times) |
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cuckoo
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Have you seen this kind of matrix?
« on: Mar 27th, 2009, 8:28am » |
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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!
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towr
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Re: Have you seen this kind of matrix?
« Reply #1 on: Mar 27th, 2009, 11:29am » |
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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.
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Wikipedia, Google, Mathworld, Integer sequence DB
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Obob
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Re: Have you seen this kind of matrix?
« Reply #2 on: Mar 27th, 2009, 2:10pm » |
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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.
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cuckoo
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Re: Have you seen this kind of matrix?
« Reply #3 on: Mar 27th, 2009, 8:24pm » |
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you are right.
Thank you all, guys!
on Mar 27th, 2009, 2:10pm, 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. |
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cuckoo
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Re: Have you seen this kind of matrix?
« Reply #4 on: Mar 27th, 2009, 8:31pm » |
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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] ?
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Obob
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Re: Have you seen this kind of matrix?
« Reply #5 on: Mar 27th, 2009, 10:18pm » |
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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.
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Eigenray
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Re: Have you seen this kind of matrix?
« Reply #6 on: Mar 27th, 2009, 11:33pm » |
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If it's a square matrix, to compute the characteristic polynomial we need only find the coefficient of tn-2. Thus
det( M - t I ) = (-t)n-2 [ t2 - S(a+b) t + S(a)S(b) - n <a,b> ],
where S(x) =  xi.
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| « Last Edit: Mar 27th, 2009, 11:36pm by Eigenray » |
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