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I have done most of the work but I struggle to put this matrix into Jordan Normal Form.

$$C=\begin{bmatrix}1 & 0 & 0 & 0&0\\1 & -1 & 0 & 0&-1 \\1 & -1 & 0 & 0&-1 \\ 0 & 0 & 0 & 0&-1\\-1&1&0&0&1\end{bmatrix}$$

The characteristic polynomial is $(x-1)x^4$, so the eigenvalues are $1,0,0,0,0$. The rank of $C$ is 3. The rank of $C^2=2$.

I know there is only $1$ Jordan block of size $1$ for the eigenvalue $1$, $J_1(1)$. But I am stuck to determine the blocks for the eigenvalue $0$. I know the block sizes must add up to $4$. Can someone help me finish finding the Jordan blocks for $0$?

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If $B$ is a Jordan block of sike $k$ for the eigenvalue $0$, then $B$ has rank $k-1$, and if $k > 1$, then $B^{2}$ has rank $k-2$. Since going from $C$ to $C^{2}$ the rank decreases by one, there must only one Jordan block of size $> 1$ for the eigenvalue $0$, and it must have size $3$ by the above, so you have a further block of size $1$ for $0$.

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  • $\begingroup$ I'm sorry, I do not understand why $B$ must have a rank $k-1$. There are $2$ blocks for eigenvalue $0$ because rank$C$ is $2$, is this correct? If so $4=3+1=2+2$, why does the rank of $C$ to $C^2$ decreasing by $1$ imply that the largest block is of size $3$? $\endgroup$ – pureundergrad May 19 '18 at 17:17
  • $\begingroup$ For each block of size $k > 1$ with respect to the eigenvalue $0$, the rank decreases by one going from the matrix $C$ to $C^{2}$. So there can be only one such block $B$. Given the contribution $1$ to the rank from the block with respect to the eigenvalue $1$, $B$ must have size $3$. $\endgroup$ – Andreas Caranti May 19 '18 at 17:22
  • $\begingroup$ The part I am struggling with is "So there can be only one such block $B$". I am trying to see what would happen if there were $2$ such blocks but I don't see what the problem is. $\endgroup$ – pureundergrad May 19 '18 at 17:33
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    $\begingroup$ If there are two blocks of size $k > 1$, then the rank goes down by $2$ when going from $C$ to $C^{2}$. $\endgroup$ – Andreas Caranti May 19 '18 at 17:38
  • $\begingroup$ @pureundersgrad I like the method at this recent question math.stackexchange.com/questions/2786604/… If you look at the question, there is a line where he indicates that he wants something in the kernel of $W^3$ for some matrix but not in kernel of $W^2$ and that is how i answered it. I am a big fan of actually working out the change of basis matrix, and its inverse, and confirming the multiplication $\endgroup$ – Will Jagy May 19 '18 at 17:39
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As at Jordan normal form (Basis)
we see here the method of finding some column vector $r_5,$ such that $C^3 r_5 = 0$ but $C^2 r_5 \neq 0.$ Once we choose that, we are forced to take $r_4 = C r_5,$ then column $r_3 = C r_4.$ This becomes an eigenvector because $Cr_3 = C^2 r_4 = C^3 r_5 = 0.$

I have picked $r_5 = (0,0,0,0,-1)^T$ and finished up. The other null vector $r_2$ can be anything in the two- dimensional subspace of genuine zero eigenvectors, so more choices are possible.

$$ \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0&-1&1&0&0 \\ 0&-1&0&1&0 \\ -1&1&0&0&0 \\ 0&-1&0&0&-1 \\ \end{array} \right) \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 1&0&0&1&0 \\ 1&1&0&1&0 \\ 1&0&1&1&0 \\ -1&0&0&-1&-1 \\ \end{array} \right) = \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0&1&0&0&0 \\ 0&0&1&0&0 \\ 0&0&0&1& 0\\ 0&0&0&0&1 \\ \end{array} \right) $$

$$ \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0&-1&1&0&0 \\ 0&-1&0&1&0 \\ -1&1&0&0&0 \\ 0&-1&0&0&-1 \\ \end{array} \right) \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 1&-1&0&0&-1 \\ 1&-1&0&0&-1 \\ 0&0&0&0&-1 \\ -1&1&0&0&1 \\ \end{array} \right) \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 1&0&0&1&0 \\ 1&1&0&1&0 \\ 1&0&1&1&0 \\ -1&0&0&-1&-1 \\ \end{array} \right) = \left( \begin{array}{c|c|ccc} 1&0&0&0&0 \\ \hline 0&0&0&0&0 \\ \hline 0&0&0&1&0 \\ 0&0&0&0& 1\\ 0&0&0&0&0 \\ \end{array} \right) $$

$$ \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 1&0&0&1&0 \\ 1&1&0&1&0 \\ 1&0&1&1&0 \\ -1&0&0&-1&-1 \\ \end{array} \right) \left( \begin{array}{c|c|ccc} 1&0&0&0&0 \\ \hline 0&0&0&0&0 \\ \hline 0&0&0&1&0 \\ 0&0&0&0& 1\\ 0&0&0&0&0 \\ \end{array} \right) \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 0&-1&1&0&0 \\ 0&-1&0&1&0 \\ -1&1&0&0&0 \\ 0&-1&0&0&-1 \\ \end{array} \right) = \left( \begin{array}{ccccc} 1&0&0&0&0 \\ 1&-1&0&0&-1 \\ 1&-1&0&0&-1 \\ 0&0&0&0&-1 \\ -1&1&0&0&1 \\ \end{array} \right) $$

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  • $\begingroup$ Thank you, but unfortunately we never got to study the basis for Jordan Normal Form in class, so I need not find the basis $\endgroup$ – pureundergrad May 19 '18 at 17:59

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