1. The problem statement, all variables and given/known data find all real values of k for which A is diagonalizable. A = [ 1 1 ] [ 0 k ] 3. The attempt at a solution let L = lamba = eigenvalue I did this: det(A - LI) = L2 - Lk - L + k so then it sorta looks like a quadratic so I did this: L2 - Lk - L + k = 0 L2 - L(k+1) + k = 0 and used the quadratic equation x = (-b +- sqrt(b2-4ac)) / 2a so then I got stuck at this point: ((k+1) +- (k-1)) / 2 and this is where I got really really stuck. I'm not sure how or what to conclude about matrix A being diagonalizable for all real values of k. or can I skip the quadratic equation method and just do this instead? (1-L)(k-L) = 0 then L = 1 and L = k so then matrix A is diagonalizable for any real number k except 0?