# 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.

Note the 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 as the dimension needed, This is how you build your relative basis.
Since $$(1,-1,1,0)\in\ker(A+I)$$ it will be zero to any power of the matrix $$(A+I)$$- so it's not of much use for us, we need 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:
$$e\rightarrow (A+I)e \rightarrow 0$$