How to find $A$ such that $A^2$ is the zero matrix? 
Let $\space t:\mathbb{R^3} \to \mathbb{R^3}$ such that $\space t(v)=0 \space \forall \space v \in \mathbb{R^3}$. The matrix form would be a $3$ by $3$, zero matrix.

If $\space t=f \circ f$, where $f:\mathbb{R^3} \to \mathbb{R^3}$ and $f\neq t$, what is the matrix form of $f$?
I know that $\space t=f \circ f= f\cdot f$ and if one let $A$ to be the matrix of $f$ and $\space T$ the matrix of $t$, then $T=A^2$. And so $A^2$ would be a zero matrix.
Is there a systematic way to solve this problem? Thanks for the hints. 
 A: Let $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear transformation satisfying $T^2 = 0$.
First of all, let us check that the kernel of $T$ has dimension at least 2.  Obviously the kernel cannot be trivial because that would mean $T$ is an isomorphism.  Let us suppose that the kernel has dimension exactly 1, generated by a vector $v \in \mathbb{R}^3$.  We can complete this to a basis $\{ v, v_1, v_2 \}$.  Since $T^2 = 0$, $T(v_1) = \lambda_1 v$ and $T(v_2) = \lambda_2 v$ for some $\lambda_1, \lambda_2 \in \mathbb{R} \setminus \{0\}$.  The problem is that $T(v_1-\frac{\lambda_1}{\lambda_2}v_2) = 0$ as well, even though $v_1-\frac{\lambda_1}{\lambda_2}v_2$ is not in the kernel!  So the kernel has dimension at least 2.
If the dimension of the kernel is 3, then the matrix for $T$ in any basis is the zero matrix.
If the dimension of the kernel is 2, then matrices for $T$ are all similar to a matrix which is zero everywhere except for a 1 at the upper-right corner.
Why is this true?  Since the kernel has dimension 2, we can find one vector $w$ which is not in the kernel.  However, $T(w)$ is in the kernel.  Now pick another vector $u$ in the kernel which is linearly independent from $T(w)$.  With respect to the basis $\{ T(w), u, w \}$, the matrix for $T$ is the special matrix we singled out.
A: If you multiply two $3 \times 3$ matrices together, the entries are the dot product of the first's rows with the second's columns. For the dot product to be zero, the rows/columns must be perpendicular. So, find a matrix where every column is perpendicular to every row (use zero rows and columns to make things easier).
