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March 30th, 2015, 08:20 PM   #1
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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
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March 30th, 2015, 09:01 PM   #2
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Quote:
Originally Posted by mophiejoe View Post
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 $-7k-32=0$ ...
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April 20th, 2015, 09:09 AM   #3
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Quote:
Originally Posted by mophiejoe View Post
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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