Given $A^2$, find all $A$ that satisfy Find the set of all matrices $A\in M_{n\times n}(\mathbb R)$ satisfying $A^2=\begin{pmatrix} 0 &1& 0 \\ 0& 0& 0 \\ 0& 0& 0 \end{pmatrix}$
I have been given this problem. Do I just need to take $A=\begin{pmatrix} a &b& c \\ d& e& f \\ g& h& i \end{pmatrix}$ and solve for the variables? or is there a better way
 A: Let $$A=\begin{pmatrix} a &b& c \\ d& e& f \\ g& h& i \end{pmatrix}$$ as OP suggests. Now clearly, 
$$A^4=0$$ and thus the minimal polynomial of $A$ divides $x^4$. However the minimal polynomial has degree $\leq 3$ and therefore the minimal polynomial divides $x^3$. Thus we have 
$$A^3=0$$ and so
$$\begin{pmatrix} a &b& c \\ d& e& f \\ g& h& i \end{pmatrix}
\begin{pmatrix} 0 &1& 0 \\ 0& 0& 0 \\ 0& 0& 0 \end{pmatrix}=0$$
And $$\begin{pmatrix} 0 &1& 0 \\ 0& 0& 0\\0& 0& 0 \end{pmatrix}
\begin{pmatrix} a &b& c \\ d& e& f \\ g& h& i \end{pmatrix}
=0$$. Which imply that 
$$A=\begin{pmatrix} 0 &b& c \\ 0& 0& 0 \\ 0& h& i \end{pmatrix}$$
 The original equation now becomes much easier to solve we get
$$A=\begin{pmatrix} 0 &b& c \\ 0& 0& 0 \\ 0& c^{-1}& 0 \end{pmatrix}
$$
A: If you try to square a generic matrix and solve the resulting system of 9 quadratic equations in 9 variables, the task appears (to me, at least) quite daunting.  However, we can simplify the task significantly by first solving the problem up to conjugation using JNF.
The first task is to use the information we have to deduce the minimal and characteristic polynomials of $A$.  Given the value of $A^2$, we know that $A^4=0$.  Therefore, the minimal polynomial of $A$ is a polynomial of degree at most $3$ which divides $x^4$, so is one of $x$, $x^2$, or $x^3$.  Since $A^2$ is nonzero, the minimal polynomial must be $m_A(x)=x^3$. Since the characteristic polynomial is of degree $3$ and is divisible by the minimal polynomial (and has the same roots, only with potentially larger multiplicities, a fact we do not need), we have that the characteristic polynomial is $p_A(x)=x^3$.
The minimal and characteristic polynomials of a matrix do not always determine a matrix up to conjugation (although in this case, they do).  In terms of the Jordan blocks, if $m_M(x)$ has $(x-a)$ as a factor with multiplicity $i$ and $p_M(x)$ has $(x-a)$ as a factor with multiplicity $j$, then the largest block of eigenvalue $a$ will have size $i$, and the sum of the sizes of the blocks with eigenvalue $a$ will be $j$.
In our particular case, all our blocks must have eigenvalue $0$, and we must have a block of size $3$, which (because we are a $3\times 3$ matrix) must be all of $A$ (up to conjugation).  Therefore, we have the following:
There exists an invertible matrix $P$ such that 
$$ A = P  \begin{pmatrix} 0 &1& 0 \\ 0& 0& 1 \\ 0& 0& 0 \end{pmatrix} P^{-1}; \qquad A^2=\begin{pmatrix} 0 &1& 0 \\ 0& 0& 0 \\ 0& 0& 0 \end{pmatrix}= P  \begin{pmatrix} 0 &0& 1 \\ 0& 0& 0 \\ 0& 0& 0 \end{pmatrix} P^{-1}.$$
At this point, one can solve for $P$ as in the solution by Rene, and then solve for $A$.
