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Find the Jordan canonical form $J$ of the matrix:$$A=\begin{pmatrix}0 &0 &0 &0 \\ 0& 0& 0& 0\\ 0 &1& 0& 0\\ 0 &1& 0& 0\end{pmatrix}.$$ Find the matrix $S$ such that $A=S^{-1}JS$.

My Try:

The characteristic polynomial of $A$ is $\lambda^4=0.$ I got 3 linearly independent eigenvectors $(1, 0,0,0),(0, 0,1,0)$ and $(0, 0,0,1)$ as columns of $S$. So $$J=\begin{pmatrix}0 &0 &0 &0 \\ 0& 0& 0& 0\\ 0 &0& 0& 1\\ 0 &0& 0& 0\end{pmatrix}.$$ How do I find the other column of $S$? Can anybody please help me? I am very poor at linear algebra.

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  • $\begingroup$ No, you didn't get it right. Note that $A$ and $J$ should have the same rank. $\endgroup$ – Omnomnomnom Oct 19 '15 at 14:52
  • $\begingroup$ @Omnomnomnom they do have the same rank(=1) $\endgroup$ – Vladimir Vaschenko Jun 13 '17 at 9:19
  • $\begingroup$ @VladimirVaschenko the comment was from before the question was edited. Originally, the answer he had was $$ J=\begin{pmatrix}0 &1 &0 &0 \\ 0& 0& 1& 0\\ 0 &0& 0& 1\\ 0 &0& 0& 0\end{pmatrix}. $$ $\endgroup$ – Omnomnomnom Jun 13 '17 at 12:09
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If $$J = \pmatrix{0 & 0 & 0 & 0\cr 0 & 0 & 0 & 0\cr0 & 0 & 0 & 1\cr0 & 0 & 0 & 0\cr}$$ the columns of $S^{-1}$ should be $u_1$, $u_2$, $u_3$, $u_4$ such that $ A u_i = S^{-1} J e_i$, where $e_i$ are the standard basis of $\mathbb R^4$. Thus $A u_1 = 0$, $A u_2 = 0$, $A u_3 = 0$, $A u_4 = u_3$. In particular $u_3$ must be in the column space of $A$ as well as in the null space of $A$. You can take $u_3 = (0,0,1,1)^T$, $u_4 = (0,1,0,0)^T$, and then e.g. $u_1 = (1,0,0,0)$, $u_2 = (0,0,0,1)$. Then

$$ S = \pmatrix{1 & 0 & 0 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 1 & 0\cr 0 & 1 & 1 & 0\cr}^{-1} = \pmatrix{1 & 0 & 0 & 0\cr 0 & 0 & -1 & 1\cr 0 & 0 & 1 & 0\cr 0 & 1 & 0 & 0\cr}$$

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  • $\begingroup$ Thanks. Just to understand, it is ok to take $u_2=(0,0,1,0)$ as well. Isn't it? $\endgroup$ – Extremal Oct 19 '15 at 17:44
  • $\begingroup$ Yes, that's OK. $\endgroup$ – Robert Israel Oct 19 '15 at 17:48

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