Finding Eigenvectors, is there always a row of $0$'s? 
Given the Matrix $$A = \left(\begin{matrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-1 & -3 & -3
\end{matrix}\right)$$
calculate the eigenvalues and the corresponding eigenvectors.

My attempt
$$P(\lambda) = \det(A-\lambda I) = -\lambda^3-3\lambda^2-3\lambda-1 =0$$ is the characteristical polynome with the solutions $\lambda_1=\lambda_2=\lambda_3 = -1$
So there is only one eigenvalue $-1$
This involves solving $(A-(-1)I)= A+I = 0$.

Is "the last row" always $0$ when calculating the eigenvectors?

Update
I've got the final answer to be $$E_{-1}=\left(\begin{matrix}1 \\ -1 \\ 1 \end{matrix}\right)$$

Given:
$$A = \left(\begin{matrix}
0 & 1 & 0 \\
0 & 0 & 1 \\
-1 & -3 & -3
\end{matrix}\right)$$
We arrive at:
$$P(\lambda) = det(A-\lambda I) = -\lambda^3-3\lambda^2-3\lambda-1 =0$$ is the characteristical polynomial with the solutions $\lambda_1=\lambda_2=\lambda_3 = -1$
Solving for the eigenvectors, we arrive at::
$$\lambda_1 = -1, v_1 = (1, -1, 1)$$
$$\lambda_2 = -1, v_2 = (2, -1, 0)$$
$$\lambda_2 = -1, v_3 = (3,-1, 0)$$
From these eigenvalues and eigenvectors, the Jordan Normal Form can be written as:
$$A = P J P^{-1} = \begin{bmatrix} 1 & 2 & 3 \\ -1 & -1 & -1 \\ 1 & 0 & 0\end{bmatrix} \cdot \begin{bmatrix} -1 & 1 & 0 \\ 0 & -1 & 1 \\ 0 & 0 & -1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 0 & 1 \\ -1 & -3 & -2 \\ 1 & 2 & 1 \end{bmatrix}$$
Regards
