Reducing the basis vectors of $Ker(A+I)^2$ using the basis vector of $Ker(A+I)$ Please consider taking a look on the example 2 in this pdf. In the attached pdf, in example 2, the author says 
Reducing the basis vectors of $Ker(A+I)^2$ using the basis vector of $Ker(A+I)$, we end up with a relative basis vector.
Can someone please explain to me (in detail ) how to obtain the relative basis vector. I am unable to follow the above step that I have mentioned. Thanks for help. Any help is appreciated.
 A: Note that $\ker(M)\subseteq\ker(M^2)$, so if you find a vector that satisfies $Mv=0$ it would also satisfy $M^2v=0$. Also note that
$$
e=\left[\begin{array}{l}{0} \\ {1} \\ {2} \\ {0}\end{array}\right]=\left[\begin{array}{r}{-2 / 3} \\ {1} \\ {0} \\ {0}\end{array}\right]+2\left[\begin{array}{c}{1 / 3} \\ {0} \\ {1} \\ {0}\end{array}\right]
$$
This is a linear combination of vectors from the kernel of $(A+I)^2$, so $e$ also satisfies $(A+I)^2e=0$, but we need another vector in our thread so we notice that $e \in \operatorname{ker}(A+I)^{2} \backslash \operatorname{ker}(A+I)$, more specifically  $(A+I)^2e=0$ but $(A+I)e\neq0$, so now we have a thread of 2 vectors, just as the dimension we needed! This is how you build your relative basis.

Edit to elaborate
Since $(1,-1,1,0)\in\ker(A+I)$ it will be zero for any power of the matrix  $(A+I)$- so unfortunately it's not of much use for us. We want a chain basis of 2 vectors! So we search for vectors from the kernel of a higher power of the matrix. Fortunately, we find 2, but all we need is one (because we already have one the one from the kernel of $A+I$). Now we take a linear combination of these vectors that when multiplied by (A+I) will produce the vector in the kernel of $(A+I)$. Take a look at the example: After obtaining $e$ we compute $(A+I)e=(-1,1-1,0)=-(1,-1,1,0)$, this a scalar-multiple of the vector from the kernel of $A+I$, and since $\ker(A+I)\subseteq\ker(A+I)^2$ we get the chain:
$$e\rightarrow (A+I)e \rightarrow 0$$
