Find the canonical Jordan form of $T\left(p(x))=p(x)^{\prime}+2 p(x)\right.$

In $$P_{2}(R),$$ let be $$T: P_{2}(R) \rightarrow P_{2}(R),$$ the operator given by $$T\left(p(x)=p(x)^{\prime}+2 p(x)\right.$$

Find the canonical Jordan form $$J$$ for $$T$$. and find a basis of vectors $$P_{2}(R)$$ where the associated matrix to $$T$$ is the matrix $$J$$.

First I find the matrix associated to T it gives me

$$M=\begin{bmatrix} 2 & 1 & 0 \\0 & 2 & 2 \\ 0 & 0 & 2\end{bmatrix}$$ in the canonical basis $${1,x,x^2}$$

Then its eigenvalues are 2 with multiplicity 3,

and then the Jordan form is $$J=\begin{bmatrix} 2 & 1 & 0 \\0 & 2 & 1 \\ 0 &0 & 2\end{bmatrix}$$

and I find the $$Ker(M-2I)$$ and it gives me $$gen={t(1,0,0)}$$

and now im stuck finding the other vector to from a basis

• Is not $M=\begin{bmatrix} 2 & 1 & 0 \\0 &2 & 2 \\ 0 & \color{red}{0} & 2\end{bmatrix}$? Commented Nov 16, 2020 at 19:56
• @aqua yes, is 0 my mistake Commented Nov 16, 2020 at 19:58
• Perhaps you want to find the vector $v$ that gives $(M-2I)v$ equal to your eigenvector? And then ... Commented Nov 16, 2020 at 19:58
• Yes I find gen={t(0,1,0)} Commented Nov 16, 2020 at 20:05
• @TedShifrin the (0,1,0) vector is another vector of the basis, but know how can I find the last vector? Commented Nov 16, 2020 at 20:06

Find a nonzero vector $$w \in \Bbb{C}^3 \setminus \ker(M-2I)^2$$. Now $$\{u,v,w\}=\{(M-2I)^2w, (M-2I)w, w\}$$ is a basis for $$\Bbb{C}^3$$ and we have $$Mu = M(M-2I)^2w = (M-2I)^3w + 2(M-2I)^2w = 2u$$ $$Mv = M(M-2I)w = (M-2I)^2w + 2(M-2I)w = u+2v$$ $$Mw = (M-2I)w+2w =v+2w$$
so it is a Jordan basis for $$M$$. Picking $$w = (0,0,1)$$ we get the basis $$\{u,v,w\} = \{(2,0,0),(0,2,0),(0,0,1)\} = \{2,2x,x^2\}.$$