Let $$A =\begin{pmatrix} 177& 548& 271& -548& -356\\ 19& 63& 14& -79& -23\\ 8& 24& 17& -20& -20\\ 42& 132& 55& -141& -76\\ 56& 176& 80& -184& -105\end{pmatrix}$$
We know that $\lambda = 3$ is an eigenvalue with algebraic multiplicity 4. Find the eigenvectors and generalized eigenvectors.
It was easy to find the eigenvectors. Finding the eigenvectors for $\lambda = 3$ involves finding the kernel of the matrix $(A -\lambda I)$, which involves finding the reduced-row echelon form. By gaussian elimination we get.
$$\left[\begin{matrix} 1 & 0 & 0 & 19/5 & -6 \\ 0 & 1 & 0 & -14/5 & 7/4 \\ 0 & 0 & 1 & 6/5 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}\right] \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix}\right] = \left[\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}\right]. $$
The eigenvector of which is $$\left[\begin{matrix} x_1 \\ x_2 \\ \frac{14}{29}x_1 + \frac{220}{203}x_2 \\ \frac{5}{29}x_1 + \frac{120}{203}x_2 \\ \frac{8}{29}x_1 + \frac{76}{203}x_2 \end{matrix}\right] = \left[\begin{matrix} x_1 \\ 0 \\ \frac{14}{29}x_1 \\ \frac{5}{29}x_1 \\ \frac{8}{29}x_1 \end{matrix}\right] + \left[\begin{matrix} 0 \\ x_2 \\ \frac{220}{203}x_2 \\ \frac{120}{203}x_2 \\ \frac{76}{203}x_2 \end{matrix}\right].$$ However, this is not enough, so we need to find the eigenvector of $(A - 3I)^2$. But using gaussian elimination gives
$$\left[\begin{matrix} 1 & 8/3 & 0 & 11/3 & -4/3 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{matrix}\right] \left[\begin{matrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{matrix}\right] = \left[\begin{matrix} 0 \\ 0 \\ 0 \\ 0 \\ 0 \end{matrix}\right].$$
Which doesn't give us the right number of generalized eigenvectors (we need two more). What went wrong, and how do I make it right?