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March 30th, 2015, 08:20 PM  #1 
Member Joined: Jan 2015 From: Colorado Posts: 32 Thanks: 0  For A to have 0 as an eigenvalue, k must be
Let A= matrix [7 8] [4 k] For A to have 0 as an eigenvalue, k must be? When I did this I thought you would just row reduce and figure out what k would be, but I was wrong. If you could help me, that would be greatly appreciated! Thanks 
March 30th, 2015, 09:01 PM  #2 
Senior Member Joined: Jan 2012 From: Erewhon Posts: 245 Thanks: 112  For $0$ to be an eigenvalue the characteristic equation must have $0$ as a root. Here the characteristic equation is $(7\lambda)(k\lambda)32=0$. Which is equivalent to the constant term being zero, so $7k32=0$ ...

April 20th, 2015, 09:09 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902  Row reduce to do what? Row reduction does NOT preserve eigenvalues you cannot use rowreduction to learn anything about eigenvalues. But what is true is that a matrix will have 0 as an eigenvalue if and only if its determinant is 0.


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eigenvalue, eigenvalues 
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