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

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

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.

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  • $\begingroup$ Let $v_1$, $v_2$ be the two eigenvectors $(1,1,0,0)$ and $(0,0,1,1)$. Try setting $Tu = v_1$ and $Tu = v_2$, and see which one gives you a unique answer and which one gives you a one-parameter space? $\endgroup$ – Andrew Tindall Aug 26 '19 at 18:38
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    $\begingroup$ Greetings! Is there any particular reason on why you deleted your previous question? (math.stackexchange.com/questions/3334999/…) $\endgroup$ – Zacky Aug 26 '19 at 18:41
  • $\begingroup$ i was told to give more insigth into where did the question came from and how far did i got into solving the problem $\endgroup$ – Tomas Sexenian Aug 26 '19 at 18:58
  • $\begingroup$ Thanks you for improving the post. For the future, however, note that you can edit your posts (there is a button below each post). Usually, it is better to improve existing posts via edits rather than deleting and reposting. Don't worry for the current case, but please keep it in mind for the future. $\endgroup$ – quid Aug 26 '19 at 19:56
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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$$\begin{bmatrix}0&0&0&0\\0&2&1&0\\0&0&2&0\\0&0&0&2\end{bmatrix}.$$

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  • $\begingroup$ Thank you so much !! , so if i got this rigth as i only need one more vector to complete the jordan basis there is only one ecuation with has a set of possible solutions? if for example i needed two vectors to complete the basis then there will be two ecuations with a possible set of solutions? Or is that just a mere coincidence of this particular case and if i only need one i just choose a random ecuation and that will do the job $\endgroup$ – Tomas Sexenian Aug 26 '19 at 19:29
  • $\begingroup$ In this case, the space has dimension $4$ and you only had $3$ linearly independent eigenvectors. So, all you needed was to solve one equation. $\endgroup$ – José Carlos Santos Aug 26 '19 at 19:36
  • $\begingroup$ Note that in this case, we got lucky that one of the eigenvectors we used in the basis for the eigenspace happened to be in the image of $T - 2 I$. In general, that won't necessarily happen, and we would need to solve $(T - 2 I)^2 v = 0$ to find more elements of the generalized eigenspace. $\endgroup$ – Daniel Schepler Aug 26 '19 at 21:41
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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 : $(2 ,-2, 0 ,0), (1, 0, 0 ,0), (0, 0 ,1, 1)$

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    $\begingroup$ Note: This answer was 'imported' from an earlier version of the question which is why it is in part redundant with the current question. $\endgroup$ – quid Aug 26 '19 at 19:51

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