1. The problem statement, all variables and given/known data 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 2. Relevant 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. 3. 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!!