Find a linear operator s.t. $A^{2}=A^{3}$ but $A^{2}\neq A$?

Question

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.

Answer 1

Hint: Find a nilpotent $2\times 2$ matrix with nilpontency order $2$.

Answer 2

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 $A^2=0\not=A$.

Answer 3

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 $$ C = \pmatrix{ 0&0&0\\ 1&0&0\\ 0&1&1 } $$

Answer 4

(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.