Find a linear operator s.t. $A^{2}=A^{3}$ but $A^{2}\neq A$? From Halmos's Finite-Dimensional Vector Spaces, question 6a section 43, the section after projections. Find a linear transformation A such that $A^{2}(1-A)=0$ but A is not idempotent (I remember A is idempotent iff it is a projection). 
I had no luck.
 A: Hint: Find a nilpotent $2\times 2$ matrix with nilpontency order $2$.
A: Take for $A$ the operator that maps basis vector 2 to basis vector 1 but maps all other basis vectors (including vector 1) to zero. Then $A^2=0\not=A$.
A: The companion matrix of the polynomial is a good counterexample to keep in mind.  In this case, considering the polynomial $x^3 - x^2$, we have
$$
C = 
\pmatrix{
0&0&0\\
1&0&0\\
0&1&1
}
$$
A: (Too long for a comment.) You have no luck because you depend on luck.
It really doesn't take much to solve the problem. If $A\ne A^2=A^3$, there must exist a vector $x$ so that when $y=Ax$ and $z=A^2x$, we have $y\ne z$ but $Az=z$. In other words, by applying $A$ repeatedly on $x$, we get the following chain of iterates:
$$
x \mapsto y \mapsto \underbrace{z}_{\ne y} \mapsto z.
$$
Now it is utterly easy to construct an example of $A$ with this chain of iterates. For instances:


*

*if we put $x=\pmatrix{1\\ 0},\ y=\pmatrix{0\\ 1}$ and $z=0$, we get ahulpke's answer (here $x$ is his "vector 2" and $y$ is his "vector 1");

*if we put $x=\pmatrix{1\\ 0\\ 0},\ y=\pmatrix{0\\ 1\\ 0}$ and $z=\pmatrix{0\\ 0\\ 1}$, we get Omnomnomnom's answer.

