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ecoist
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 Find the characteristic polynomial   « on: Feb 22nd, 2008, 6:45pm » Quote Modify

Let A=(aij) be a 2009x2009 matrix with aij=2 if i+j=2010, and aij=1 otherwise.  Find the characteristic polynomial, CA(x), of A.
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Obob
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 Re: Find the characteristic polynomial   « Reply #1 on: Feb 22nd, 2008, 7:48pm » Quote Modify

It is possible to write down 2009 linearly independent eigenvectors for A.  There are 1004 with eigenvalue 1, 1004 with eigenvalue -1, and 1 with eigenvalue 2010.  So CA(x)=(x-1)1004(x+1)1004(x-2010).
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Eigenray
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 Re: Find the characteristic polynomial   « Reply #2 on: Feb 22nd, 2008, 8:49pm » Quote Modify

Look at A2.
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Icarus
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 Re: Find the characteristic polynomial   « Reply #3 on: Feb 22nd, 2008, 9:44pm » Quote Modify

Nice, but do you have a good argument for showing that the eigenvalues of A are equally split between -1 and 1, or even that the final eigenvalue is 2010, not -2010?

I'm not seeing it myself (other than the long way that leads directly there), but then that doesn't say much.
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Eigenray
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 Re: Find the characteristic polynomial   « Reply #4 on: Feb 22nd, 2008, 10:10pm » Quote Modify

...and then look at tr(A).
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Icarus
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 Re: Find the characteristic polynomial   « Reply #5 on: Feb 23rd, 2008, 7:24am » Quote Modify

Okay, even I can see it now.
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"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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