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.