I need to find values of k so that the matrix is not diagonalizable over $\Bbb C$.
$$\begin{bmatrix}0&1&0\\0&0&1\\0&-k-7&k+8\end{bmatrix}.$$
So to do this, I found the eigenvalues which were $λ = 1, λ = 0$, and $λ = k + 7$. Then I found the eigenvectors, so $λ =1$ has eigenvectors $(1, 1, 1), λ = 0$ has eigenvectors $(1, 0, 0)$, and $λ = k +7$ has eigenvectors $(1/(7+k)^2, 1/(7+k), 1)$. Therefore, I was able to conclude that if $k = -7$ the matrix is not diagonalizable since the number of eigenvectors is different from the number of matrix rows. I also concluded that $k = 0$ and $k = 1$ will make the matrix diagonalizable. But I can't seem to find other values of $k$ that will make the matrix not diagonalizable. Can someone give me a hint? I know that there are supposed to be two $k$ values.
If $k\ne-7$ and $k\ne-6$, the matrix is diagonalizable, since it has three distinct eigenvalues. And if $k=-7$ or $k=-6$, then the matrix only has two eigenvalues. Since the dimension of the eigenspace corresponding to those eigenvalues is $1$, the matrix is not diagonalizable in those cases.