- #1
lubricarret
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Homework Statement
1. Let A =
[-8 k
0 -8]
Then A is diagonalizable exactly for the following values of k
2. Let B =
[-8 k
0 1]
Then B is diagonalizable exactly for the following values of k
Homework Equations
-Equations for eigenvalues, eigenvectors... and D=PA(P^-1)
-A matrix is invertible if and only if its determinant does not equal 0
-If matrix P created from the eigenvectors [X_1, X_2] is invertible, then matrix A is diagonalizable.
The Attempt at a Solution
I seem to only get half of this answer correct, and am not sure why. I tried solving for the eigenvectors and then checking if the matrix P (from the eigenvectors) has a determinant = to zero.
For the first part, I got the eigenvalue -8 with an algebraic multiplicity of 2; then when solving for the eigenvectors, I get:
[0 -k
0 0]
and the eigenvector t[-k,0] and the matrix P =
[-k -k
0 0]
Therefore, I say that the matrix A can never be diagonalizable for any value of k, as the number of parameters will always be less than the algebraic multiplicity, and the determinant will always be = 0, regardless of the k value.
For the second part, I get eigenvalues -8 and 1.
I then get the eigenvectors t[1,0] and t[(-k/9),1], and the matrix P =
[1 (-k/9)
0 1]
Therefore, I said this matrix will be diagonalizable for all values of k, since the determinant will always equal 1, and is therefore always diagonalizable, regardless of the k value.
Any ideas on what I am doing wrong; and if my logic is messed up?
Thanks!