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Topic: eigenvalues multiplicity (Read 2327 times) |
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MonicaMath
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eigenvalues multiplicity
« on: Oct 19th, 2009, 6:11pm » |
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I wanna prove that, for nxn matrix A:
if rank(A)=rank(A^2) then:
the algebraic multiplicity and geometric multiplicity of the zero eigenvalue are equal.
can we prove this using Jordan canonical forms?
I started by:
if rank(A) = rank (A^2), then null(A) = null(A^2)
meaning, the dimensions of the kernels of A and A^2 are equivalent,
i.e. geometric multiplicities of 0 are equal in both A and A^2,
but what about algebraic multiplicity
could anyone help me ?!
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Obob
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Re: eigenvalues multiplicity
« Reply #1 on: Oct 19th, 2009, 7:14pm » |
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Yes, you can use Jordan canonical form. If you have a matrix in Jordan canonical form, then the algebraic multiplicity of the zero eigenvalue is just the number of zeroes along the diagonal. The geometric multiplicity of the zero eigenvalue is the number of Jordan blocks with eigenvalue zero. So to show the algebraic multiplicity and geometric multiplicity are the same, you need to know that all the Jordan blocks with eigenvalue zero have size 1x1. But you can check directly that if A has a Jordan block M with eigenvalue zero and size bigger than 1x1, then M^2 has rank smaller than the rank of M, and hence also the rank of A^2 (which is the sum of the ranks of its blocks; you can calculate A^2 by multiplying corresponding blocks) is smaller than the rank of A. Thus no block with eigenvalue zero can have size bigger than 1x1.
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| « Last Edit: Oct 19th, 2009, 7:15pm by Obob » |
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MonicaMath
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Re: eigenvalues multiplicity
« Reply #2 on: Oct 19th, 2009, 8:30pm » |
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Thank you for the hint, but I didn't understant why is
" So to show the algebraic multiplicity and geometric multiplicity are the same, you need to know that all the Jordan blocks with eigenvalue zero have size 1x1"
I know that algebraic multiplicity of zero is the sum of all dimentions of the jordan blocks corresponding to the zero eigenvalue, and the geometric multiplicity is the number of jordan blocks for the zero eigenvalue.
could you explain that ?
Thank you
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| « Last Edit: Oct 19th, 2009, 9:00pm by MonicaMath » |
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Obob
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Re: eigenvalues multiplicity
« Reply #3 on: Oct 19th, 2009, 9:03pm » |
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If there is a Jordan block of size two for the zero eigenvalue, then the number of blocks is strictly smaller than the sum of the dimensions of the blocks, and hence the geometric multiplicity is smaller than the algebraic multiplicity. Conversely, if all the Jordan blocks (corresponding to zero) are 1x1, then the number of blocks is equal to the sum of the dimensions of all the blocks.
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MonicaMath
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Re: eigenvalues multiplicity
« Reply #4 on: Oct 19th, 2009, 9:06pm » |
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I got it.. thank you
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