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 $A$?
 A: 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.
A: 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 $1$.
