Find a Jordan basis for the linear operator $T$ Find a possible Jordan basis for the linear operator $T$ such that:


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*$T(x, y, z, t) = (2y, −2x + 4y, z + t, z + t)$

Is there an specific method to find a Jordan basis? Since I'm teaching myself I'm not aware of any sort of procedure into finding one.

However I managed to calculate the characteristic polynomial of $T$,  $p(t)= t(t-2)(t-2)(t-2)$ concluding that the proper sub-space associated to $t=0$ is generated by $[(0,0,1,-1)]$ and the one associated to $t=2$ is generated by $[(1,1,0,0),(0,0,1,1)]$ which is why $T$ is non-diagonizable.

That's as far as I could get by myself, so I would appreciate any kind of explanation on how to find a Jordan basis or simply solving this exercise.
I'm not used to speaking English so feel free to edit the question as I know the idea can be transmitted in a much better way.
 A: The characteristic polynomial of the operator is $p(t)=t\cdot (t-2)^3$. Using the matrix $A$ associated to the operator you can easily find that a basis for the set of the solution of the system $Ax=0$ is $(0, 0 ,-1 ,1)$. 
In the same way you can find a Jordan basis for $V_{2}$ where $V_{2}$ denotes the eigenspace of eigenvalue 2, which is : $(2 ,-2, 0 ,0), (1, 0, 0 ,0), (0, 0 ,1, 1)$
A: What you did is fine. So, you know that $T(0,0,1,-1)=0$, that $T(1,1,0,0)=2(1,1,0,0)$, and that $T(0,0,1,1)=2(0,0,1,1)$. And there is no eigenvector linearly independent of these ones. ince the characteristic polynomial of $T$ is $\lambda(\lambda-2)^3$, you know that the Jordan normal form of $T$ shall have one $0$ and three $2$'s in its main diagonal. Now, consider the equations$$T(x,y,z,t)=2(x,y,z,t)+(1,1,0,0)\text{ and }T(x,y,z,t)=2(x,y,z,t)+(0,0,1,1).$$It turns out that the second equation has no solutions, bt the first one has. For instance, $\left(-\frac12,0,0,0\right)$ is a solution. So, take the babasis$$B=\left\{(0,0,1,-1),(0,0,1,1),\left(-\frac12,0,0,0\right),(1,1,0,0)\right\}.$$The matrix of $T$ with respect to $B$ is$$\begin{bmatrix}0&0&0&0\\0&2&1&0\\0&0&2&0\\0&0&0&2\end{bmatrix}.$$
