Finding eigenvalues and eigenvectors of a matrix

  • #26
shmoe
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Treadstone 71 said:
Ok I redid my computions 4 times and I found that I left out a "-" sign somewhere. Now, however, I only have 2 eigenvalues, since I got [tex](2+L)^2(31+17L)=0[/tex], where L is the eigenvalue. I'm supposed to show that the matrix is diagonalizable, but since I only have 2 distinct eigenvalues, this doesn't work out.

The matrix in post #17 has 3 distinct eigenvalues, you might want to check your work again.

In any case, it's not necessary to have 3 distinct eigenvalues for a 3x3 matrix to be diagonalizable. You've made no mention of the corresponding eigenspaces, or any eigenvectors at all for that matter. Can you find an eigenvector corresponding to L=-31/17? This should tell you something was off in your calculation.
 
  • #27
matt grime
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Treadstone 71 said:
I'm supposed to show that the matrix is diagonalizable, but since I only have 2 distinct eigenvalues, this doesn't work out.


Consider the obviously diagonalizable identity matrix with 1s on the diagonal and zeroes elsewhere. This has only one eigenvalue, do you think that means that the identity matrix is not a diagonalizable matrix?
 

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