4 transformation matrices to jordan form

[nhrock3]

4)
$$A=\left(\begin{array}{cc}4 & -4\\1 & 0\end{array}\right)$$
find the jordan form and the transformation matrices P to this jordan
form.
the caracteristic and minimal polinomial is $$P(t)=M(t)=(t-2)^{2}$$
so the jordan form is $$J_{A}=\left(\begin{array}{cc}2 & 1\\0 & 2\end{array}\right)$$.
my prof taught a method of finding the P
he gave an example like if $$J_{B}=\left(\begin{array}{cc}0 & 1\\ 0 & 0\end{array}\right)$$
then we use chain method ,for the first column $$Tv_{1}=0$$ for the
second its $$Tv_{2}=v_{1}$$
so $$v_{1}\in kerT$$ $$v_{2}\in kerT^{2}$$
i have two question:
regarding the example of the prof $$kerT^{2}$$ is $$R^{2}$$ what vector
to pick for $$v_{2}$$?
regarding my original example i have for the first coulumn of the
jordan form $$Tv_{1}=2v_{1}$$ for the second $$Tv_{2}=v_{1}+2v_{2}$$
so i can't assosiate v\_1 and v\_2 with kernel of T
i know that i can assosiate v1 with Ker(T-2I)
but then i lose the chain method my prof taught.
what to do?

 

[lippyka]

For the first example, you can pick any vector for v2 since KerT^2 = R^2. For the second example, you can still use the chain method. The idea is to find vectors such that they satisfy the equations Tv1 = 2v1 and Tv2 = v1 + 2v2. In other words, you want to find a basis of Ker(T-2I) and Ker(T-2I) + Span{v1}. Once you have found these two vectors, you can form the matrix P by putting them as columns. This matrix P will give you the transformation from A to its Jordan form.