My Math Forum Eigenvalue

 Linear Algebra Linear Algebra Math Forum

 April 15th, 2010, 04:40 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, 05:27 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 => |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.
 April 17th, 2010, 02:33 PM #3 Senior Member   Joined: Sep 2009 Posts: 115 Thanks: 0 Re: Eigenvalue Oh ok that makes sense. Thank you very much.

 Tags eigenvalue

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post johnnyb Linear Algebra 3 September 17th, 2013 11:42 AM subhajit028 Linear Algebra 3 September 16th, 2013 12:00 PM colt Linear Algebra 3 April 24th, 2013 10:46 AM tinynerdi Linear Algebra 3 April 5th, 2010 12:56 AM Ackoo Linear Algebra 1 May 7th, 2008 07:19 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top