Characteristic polynomial of a matrix $A$ is $x^7$ and $\operatorname{rank}(A)=4$ and $\operatorname{rank}(A^2)=1$. Classify $A$. I know how to do this problem the 'long' way. I was wondering if there
was an easier, less computationally cumbersome way to do this.
Here is the question: Let $A$ be a square matrix over $\mathbb R$. Suppose that
the characteristic polynomial of $A$ is $x^7$ and $\operatorname{rank}(A)=4$ and $\operatorname{rank}(A^2)=1$. Classify all such matrices up to similarity.
My strategy is to list all the possible minimal polynomials and then for each case list the possible invariant factors (using Caley Hamilton) and then look at the RCF 
in each case and see if it satisfies the two other conditions. But this seems not a very quick way to do this. Is there some other way (for example, one that circumvents listing the minimal polynomials) to do this?. 
Also, for typical problems like this is there a standard way to proceed?. Is it easier to use, for example Jordan forms rather than rational canonical forms?.
As always all your help is greatly appreciated.
 A: Jordan form is definitely the way to proceed.  We know the matrix is $7\times 7$ and nilpotent by its characteristic polynomial, so all of the Jordan blocks have eigenvalue $0$.  This means the matrix is similar to a matrix of all zeroes, except for some $1$'s on the superdiagonal.  Since the square of the matrix has rank $1$, we must have exactly one Jordan block of size $3$, and the rest must be of size $1$ or $2$.  The rank $4$ condition says we must have two other blocks of size $2$.
In other words, all matrices you've described are similar to the following matrix:
$$\begin{pmatrix}0&1&0&0&0&0&0\\0&0&1&0&0&0&0\\0&0&0&0&0&0&0\\0&0&0&0&1&0&0\\0&0&0&0&0&0&0\\0&0&0&0&0&0&1\\0&0&0&0&0&0&0\\\end{pmatrix}$$
In general, when classifying matrices up to similarlity, Jordan form is very helpful.  Over an algebraically closed field, all matrices can be put in Jordan form via a similarity transformation.  In this case, over $\mathbb{R}$, all of the eigenvalues of a nilpotent matrix are in $\mathbb{R}$ (they're all zero), so it still has a Jordan form.
A: I would do this in my head as follows. Having a power of $x$ as characterisitic polynomial means $A$ is nilpotent, and we need to find its Jordan type (multiset of sizes of the Jordan blocks). Think of each Jordan block as a row of boxes, arrange then vertically in decreasing order of size, with the leftmost box aligned; this gives the Young diagram of the Jordan type. The number of boxes is the rank of$~A^0$, which is$~7$, the degree of the characterisitic polynomial. Each application of $A$ chops of the leftmost column of the diagram. Saying $\def\rk{\operatorname{rank}}\rk A=4$ means $4$ boxes remain after doing this once, so the first column had $7-4=3$ boxes. Saying $\rk A^2=1$ means one box remains after doing it twice, so the second column had $4-1=3$ boxes. The remaining box must be in the third column (and the first row). This makes the Jordan type $(3,2,2)$, and $A$ is similar to the Jordan normal forma matrix with those sizes for its blocks.
A: The characteristic polynomial of $A$ is $x^7$. So, $A$ is $7\times 7$ and all its eigenvalues are zero. Now consider the Jordan form of $A$. Let $A$ has $x_n$ Jordan blocks of size $n$, i.e. the Jordan block $J_n(0)$ occurs $x_n$ times. Then the problem boils down to finding all nonnegative integer solutions for the system of equations
\begin{align*}
x_1+2x_2+3x_3+4x_4+\ldots&=7,\\
     x_2+2x_3+3x_4+\ldots&=4,\\
          x_3+2x_4+\ldots&=1.
\end{align*}
The solution is obviously given by $x_3=1,\,x_2=2$ and all other $x_i$s are zero. Therefore $A$ is similar to $J_3(0)\oplus J_2(0)\oplus J_2(0)$.
