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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
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Quote:
 Originally Posted by mophiejoe Let A= matrix [-7 -8] [-4 k] For A to have 0 as an eigenvalue, k must be?
For $0$ to be an eigen-value 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 $-7k-32=0$ ...

April 20th, 2015, 09:09 AM   #3
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Quote:
 Originally Posted by mophiejoe 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
Row reduce to do what? Row reduction does NOT preserve eigenvalues- you cannot use row-reduction 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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