April 15th, 2010, 05:40 PM  #1 
Senior Member Joined: Sep 2009 Posts: 115 Thanks: 0  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?

April 17th, 2010, 06:27 AM  #2 
Newbie Joined: Jan 2010 Posts: 21 Thanks: 0  Re: Eigenvalue
Well, it was under your nose all this time. As you've mentioned : A=0 => A0*I=0 => 0 is eigenvalue of A. There are more ways to prove it, but this one is the easiest one, in my opinion. 
April 17th, 2010, 03:33 PM  #3 
Senior Member Joined: Sep 2009 Posts: 115 Thanks: 0  Re: Eigenvalue
Oh ok that makes sense. Thank you very much.


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