# Help with jordan normal form exercises

Let $A$ be a $7\times 7$ matrix such that $(A-I)^3=0$ and $(A-I)^2$ has rank $2$. How can we find the Jordan normal form of $A$?

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Since $(A-I)^3=0$, The only eigenvalue of $A$ is 1, and the Jordan blocks are of size no bigger than 3. Since $(A-I)^2$ has rank 2, There must be some Jordan block of rank bigger than 2. But if $B$ is a jordan block of size 3, $(B-I)^2$ has rank one. Hence we must have 2 Jordan blocks of size 3. Since $A$ is $7\times 7$, there is one additional Jordan block, of size 1. $$\begin{pmatrix} 1 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 1 & 0 & 0\\0 & 0 & 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0 & 0 & 1 & 0 \\0 & 0 & 0 & 0 & 0 & 0 & 1 \end{pmatrix}$$

If you don't see how I arrived at this solution, try squaring and cubing the above matrix to see what happens,and it should become more clear.

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Thanks Frankel and Jordan. By using your answer I try to solve this question. A is 4x4 matrix with complex coefficient such that $(A-3I)^2=0$ possible jordan forms of A. The only eigenvalue 3 and the size of Jordan blocks are not bigger than 2 since $(A-3I)^2=0.$ $\begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 3 & \end{pmatrix}$ $\begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 1 & 0 \\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 3 & \end{pmatrix}$ $\begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 1& 0 \\ 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 3 & \end{pmatrix}$ Is it right? – the code Feb 25 '12 at 3:53
What you said, in words, is correct. Of the matricies you wrote, only the first is a possible solution. The second has a Jordan block of size 3, and the third is a Jordan block of size 4. There are other solutions: you could 4 Jordan blocks of size 1 (ie a diagonal matrix), or one of size 2 and two of size 1. – Brett Frankel Feb 25 '12 at 4:01
$\begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 1\\ 0 & 0 & 0 & 3 & \end{pmatrix}$ $\begin{pmatrix} 3 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 \\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 3 & \end{pmatrix}$ $\begin{pmatrix} 3 & 0 & 0 & 0 \\ 0 & 3 & 0& 0 \\ 0 & 0 & 3 & 0\\ 0 & 0 & 0 & 3 & \end{pmatrix}$ – the code Feb 25 '12 at 4:49
Thank you so much Frankel I think I get it. Now it is completed right. – the code Feb 25 '12 at 4:51
Yes, this is correct. There are two other solutions, but they are essentially the same as your second solution, just with the Jordan blocks in different orders. – Brett Frankel Feb 25 '12 at 4:58

All of $A$'s eignevalues are $1$, since $A$'s minimal polynomial divides $(X-1)^3$. So the Jordan blocks have $1$s down the diagonal. The blocks most be less than or equal to $3$ in size, or else $(A-I)^3$ would not be $0$. We could not have all of the blocks being of size $\leq2$, or else $(A-I)^2$ would already be $0$ and have $0$ rank. And each block of size $3$ can only contribute $1$ to the rank of $(A-I)^2$. So we must have two blocks of size $3$ and one of size $1$.

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