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The question asked to find the Jordan Canonical Form and Jordan Basis of $\begin{bmatrix}1 & 1 & 0 & -1\\0 & 1 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\end{bmatrix}$, and after finding the characteristic polynomial, I used its eigenvalues and their multiplicities to find the JCF.

I found the JCF to be $\begin{bmatrix}0 & 0 & 0 & 0\\0 & 1 & 1 & 0\\0&0&1&1\\ 0 & 0 & 0 & 1\end{bmatrix}$, but I am unsure of how to find a Jordan Basis for this.

Intuitively, I'd think the Jordan Basis for this would be the vectors that make up the three linearly independent columns, but my textbook doesn't really explicitly state what the basis is when they define the JCF. Is my intuition correct or am I way off? If so how can I find the Jordan Basis for this JCF?

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  • $\begingroup$ Well you know The Jordan Canonical Form is written in some basis. You have the coordinates, the eigenvectors and the original Matrix... $\endgroup$ – TheNicouU Sep 30 '18 at 20:16
  • $\begingroup$ I think I get what you're hinting at but I'm not 100% sure. Am I supposed to apply the JCF to some basis vectors to find their corresponding basis vectors in the Jordan Basis? If so how do I find those original basis vectors? $\endgroup$ – bloominotti Sep 30 '18 at 20:32
  • $\begingroup$ I think you are a little confused. The JCF is written in some basis, which is formed by vectors you already know(eigenvectors) and other that you do not know. Now if you want to find the other vectors you just need to realise that in the JCF you have coordinates, each corresponding to a vector. For example, consider the third column of the JCF you have. You know that (lets call your original matrix $A$): $Av_3 = 1v_2 + 1v3$ Thats a linear system. You know $v_2$ (which should be the eigenvector corresponding to the eigenvalue 1). Now you just have to solve that system for a generic vector $v_3$ $\endgroup$ – TheNicouU Sep 30 '18 at 20:47
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let your matrix be $A$ and name $B = A - I.$ the characteristic polynomial says $A B^3 = 0.$ This is also the minimal polynomial.

The method I like is to find a column vector $w$ with $B^3 w = 0$ but $B^2 w \neq 0.$ I like zeros and ones, so I am choosing $$ w = \left( \begin{array}{r} 0 \\ 0 \\ 0 \\ 1 \\ \end{array} \right) $$ This is going to be the far right column, number 4. Next $v = B w$ with $$ v = \left( \begin{array}{r} -1 \\ 1 \\ 0 \\ 0 \\ \end{array} \right) $$ We reach a genuine eigenvector finally with $u = Bv$ $$ u = \left( \begin{array}{r} 1 \\ 0 \\ 0 \\ 0 \\ \end{array} \right) $$ because $Bu = B^2 v = B^3 w = 0,$ so $Au = u.$ The 0 eigenvector is the first column. Call the whole thing $R$ $$ R = \left( \begin{array}{rrrr} 0&1&-1&0 \\ 0&0&1&0 \\ 1&0&0&0 \\ 0&0&0&1 \\ \end{array} \right) $$ Next calculate $\det R = 1$ and $$ R^{-1} = \left( \begin{array}{rrrr} 0&0&1&0 \\ 1&1&0&0 \\ 0&1&0&0 \\ 0&0&0&1 \\ \end{array} \right) $$ finally $J = R^{-1} A R$

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  • $\begingroup$ I think I understand it now, my professor did something very similar in starting with a vector in B^3 but not B^2. Just to make sure I understand this, in this case the basis would be the 0 eigenvector, u, v, and w? Also, would the columns of R^-1 also form a different basis for the same JCF? (Sorry my formatting is terrible, I am new here and for some reason the format preview box only appeared for the original question) $\endgroup$ – bloominotti Sep 30 '18 at 21:42

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