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April 15th, 2010, 05:40 PM   #1
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Eigenvalue

How do you prove if A is an n by n singular matrix, then 0 must be an eigenvalue of A. It make sense since we know that det A=0. To find the eigenvalues, it is det (A - hI)=0..[h being the eigenvalues] I am stuck right now. I know that when you do the det (A - hI) you will end up with xh+yh^2+..... so you can pull out an h(x+yh+....)=0 so h=0. Is there a better way to formulate this proof?
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April 17th, 2010, 06:27 AM   #2
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Re: Eigenvalue

Well, it was under your nose all this time. As you've mentioned :
|A|=0 => |A-0*I|=0 => 0 is eigenvalue of A.
There are more ways to prove it, but this one is the easiest one, in my opinion.
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April 17th, 2010, 03:33 PM   #3
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Re: Eigenvalue

Oh ok that makes sense. Thank you very much.
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