Find the eigenvalues of...

Question

My characteristic equation starts off:

$$\lambda(\lambda(\lambda-3k)+3k^2)-k^3=0$$ Once expanded I get: $$\lambda^3-3\lambda^2k+3\lambda k^2-k^3=0$$

Where do I go from here?

Answer 1

Factor! $$ 0 = \lambda^3 - 3\lambda^2 k + 3\lambda k^2 - k^3 = (\lambda - k)^3 $$ so $\lambda = k$ is the only root (eigenvalue).