How did we find the generalized eigenvectors corresponding to the eigenvalue 2 here pg.305 in Golan. The example is given below:



How did we find the generalized eigenvectors corresponding to the eigenvalue 2 here?
I watched this video https://www.youtube.com/watch?v=msFp3vOYIoA and I followed the procedure mentioned in it, which say take the eigenvector you calculated first and it is now no longer homogenous equation that you want to find its solution but AX = the eigenvector that you found in the first step. but this procedure did not give me the vectors mentioned in the picture. Could anyone help me please?  
 A: If you're good with generalized eigenvectors as a concept, skip this. Recall for an eigenvector $Av=\lambda v$ so $(A-\lambda I)v=0$. For generalized eigenvectors we satisfy $(A-\lambda I)^kv=0$, so we can see that an eigenvector can be generalized by solving for $(A-\lambda I)u=v$ where $v$ is a generalized eigenvector of degree $k$ and $u$ of degree $k+1$. Then $(A-\lambda I)^{k+1}u=(A-\lambda I)^k(A-\lambda I)u=(A-\lambda I)^kv=0$. 
To the computation. Did you check to subtract $2I$ from the original matrix? 
$$(A-2I)v=
\begin{bmatrix} 
0 & -2 & 1 & 1 \\
0 & -1 & 1 & 1 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}v=0$$ 
gives us our first eigenvector of $$
\begin{bmatrix} 
1 \\
0 \\
0 \\
0 \\
\end{bmatrix}$$ by inspection. A lone zero column does us justice.
Then we want to solve $(A-\lambda I)u=v$ or: $$\begin{bmatrix} 
0 & -2 & 1 & 1 \\
0 & -1 & 1 & 1 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{bmatrix} \begin{bmatrix} 
a \\
b \\
c \\
d \\
\end{bmatrix} = \begin{bmatrix} 
1 \\
0 \\
0 \\
0 \\
\end{bmatrix}$$
Note to get the third zero, we need that $\begin{bmatrix} 
0 & 0 & 0 & 1 \\
\end{bmatrix}\begin{bmatrix} 
a \\
b \\
c \\
d \\
\end{bmatrix}=\begin{bmatrix} 
0 \\
\end{bmatrix}$ so it must be that $d=0$. Then the second row must satisfy $\begin{bmatrix} 
0 & -1 & 1 & 0 \\
\end{bmatrix}\begin{bmatrix} 
a \\
b \\
c \\
0 \\
\end{bmatrix}=\begin{bmatrix} 
0 \\
\end{bmatrix}$ giving $-b+c=0$ so $b=c$. Then we have from the first row $\begin{bmatrix} 
0 & -2 & 1 & 1 \\
\end{bmatrix}\begin{bmatrix} 
a \\
b \\
b \\
0 \\
\end{bmatrix}=\begin{bmatrix} 
1 \\
\end{bmatrix}$ so $-b=1$ gives $b=c=-1$ so $\begin{bmatrix} 
a \\
b \\
c \\
d \\
\end{bmatrix}=\begin{bmatrix} 
0 \\
-1 \\
-1 \\
0 \\
\end{bmatrix}$
One more time:
$$\begin{bmatrix} 
0 & -2 & 1 & 1 \\
0 & -1 & 1 & 1 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{bmatrix} \begin{bmatrix} 
e \\
f \\
g \\
h \\
\end{bmatrix} = \begin{bmatrix} 
0 \\
-1 \\
-1 \\
0 \\
\end{bmatrix}$$
Start with the third row, so $\begin{bmatrix} 
0 & 0 & 0 & 1 \\
\end{bmatrix}\begin{bmatrix} 
e \\
f \\
g \\
h \\
\end{bmatrix}=\begin{bmatrix} 
-1 \\
\end{bmatrix}$ so $h=-1$. Then from the 2nd row, $\begin{bmatrix} 
0 & -1 & 1 & 1 \\
\end{bmatrix}\begin{bmatrix} 
e \\
f \\
g \\
-1 \\
\end{bmatrix}=\begin{bmatrix} 
-1 \\
\end{bmatrix}$ so $-f+g+h=-f+g-1=-1$, so $-f+g=0$ or $f=g$.
By the first row, this implies that $\begin{bmatrix} 
0 & -2 & 1 & 1 \\
\end{bmatrix}\begin{bmatrix} 
e \\
f \\
f \\
-1 \\
\end{bmatrix}=\begin{bmatrix} 
0 \\
\end{bmatrix}$ so $-2f+f+h=-f-1=0$, or $f=g=-1$.
So we get $\begin{bmatrix} 
e \\
f \\
g \\
h \\
\end{bmatrix}=\begin{bmatrix} 
0 \\
-1 \\
-1 \\
-1 \\
\end{bmatrix}$.
