My Math Forum Not finding one of the eigenvalue

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 April 23rd, 2013, 10:46 PM #1 Newbie   Joined: Apr 2013 Posts: 10 Thanks: 0 Not finding one of the eigenvalue Hi, I have this matrix $4 \times 4~matrix $\left( \begin{array}{ccc} 2 & 0 & 0 & 0 \\ 3 & -4 & 0 & -2 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & 0 & 2 \\ \end{array} \right)$$ As the title says, I try to find the eigenvalues for it. I do this by doing: L2 -> L2 -3/2L1 and L4 -> l4 - 1/2L1 This way I get a triangular matrix like: $4 \times 4~matrix $\left( \begin{array}{ccc} 2-\lambda & 0 & 0 & 0 \\ 0 & -4-\lambda & 0 & -2 \\ 0 & 0 & 1-\lambda & 0 \\ 0 & 0 & 0 & 2-\lambda \\ \end{array} \right)$$ Then gets a determinant of $(2-\lambda) * (-4-\lambda) * (1-\lambda) * (2-\lambda)$ with leads me to eigenvalues of {-2,4,-1} However, when I look the answer, I see that, beyond these values, -3 is also a eigenvalue. So what I am missing? Thanks for the help
 April 24th, 2013, 07:30 AM #2 Newbie   Joined: Apr 2013 Posts: 11 Thanks: 0 Re: Not finding one of the eigenvalue I checked it with MATLAB. I found out that eigen values are -4,2 and 1 So you are correct other than the signs are just opposite
 April 24th, 2013, 10:33 AM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Not finding one of the eigenvalue You seem to be saying that you first row-reduced the matrix, then found the eigenvalues of the row-reduced matrix. You cannot do that- the eigenvalues of the row reduced matrix are not necessarily the same as the eigenvalues of the original matrix. What you can can do is expand the eigenvalue determinant by "rows" and 'columns". Expanding on the first row, $\left|\begin{array}{cccc}2- \lambda & 0 & 0 & 0 \\ 3 & -4- \lambda & 0 & -2 \\ 0 & 0 & 1- \lambda & 0 \\ 1 & 0 & 0 & 2-\lambda\end{array}\right|$ we have $(2- \lambda)\left|\begin{array}{ccc}-4- \lambda & 0 & -2 \\ 0 & 1- \lambda -2 \\ 0 & 0 & 2- \lambda\end{array}\right|$ That is now a "triangular" array so its determnant is the product of numbers on the diagonal: $(2- \lambda)(-4- \lambda)(1- \lambda)(2- \lambda)$. 2 is a double eigenvalue, 1 and -4 are the other eigenvalues.
 April 24th, 2013, 10:46 AM #4 Newbie   Joined: Apr 2013 Posts: 10 Thanks: 0 Re: Not finding one of the eigenvalue Hi, HallsofIvy. There is something missing in your latex commands, so I didn't understand what you did in the expansion of the eigenvalue determinant. Can you try to do it again? Thanks to both

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