I am trying to calculate a Jordan basis for the matrix A. $$A=\begin{pmatrix}1&1\\-1&3\end{pmatrix}$$ $$\chi(A)=(\lambda-2)^2 \to \lambda =2 , \alpha =2$$ $$V(2)=ker(A-2I)= span\begin{pmatrix}1\\1\end{pmatrix}$$ In my understanding to find the other basis vector, one chooses $v=\begin{pmatrix}1\\1\end{pmatrix}$ and computes $Bv \ne 0 , B=A-2I$, which will give the other vector in the Jordan basis of A. However I get the zero matrix in this case, I also tried to compute $ker(A-2I)^2$ but $(A-2I)^2$ is also the zero matrix.
Help would be greatly appreciated.
Pick a vector $w$ which is not in $\ker(A-2I)$, say $w = \pmatrix{0 \\ 1}$ and define $v = (A-2I)w = \pmatrix{1 \\ 1}$.
Then $v \in \ker(A-2I)$ so $$Av = 2v, \quad Aw = (A-2I)w +2w = v + 2w$$ which means that with respect to the basis $\{v,w\}$ the operator $A$ has matrix $$\pmatrix{2 & 1 \\ 0 & 2}.$$