Simple matrix equation I believe I'm missing an important concept and I need your help.
I have the following question:
"If $A^2 - A = 0$ then $A = 0$ or $A = I$"
I know that the answer is FALSE (only because someone told me) but when I try to find out a concrete matrix which satisfies this equation (which isn't $0$ or $I$) I fail.
Can you please give me a direction to find a concrete matrix? What is the idea behind this question?
Guy
 A: Certainly, it is true that $A^2=A$ if either $A=0$ or $A=I$. So it should hold if you join the two together: $$\begin{pmatrix}1&0\\0&0\end{pmatrix}.$$
That is, a very small $I$ ($=1$) in the upper left, and an equally small $0$ in the lower right. The off-diagonal zeroes keep them from interfering with each other.
A: You can take, for example, $\text{diag}(1,0)$.
A: $$A=\begin{pmatrix}
1 & 1 \\ 0 & 0
\end{pmatrix}$$
Read about Idempotent Matrices.
A: If for a polynomial $p$ and a matrix $A$ you have $p(A)=0$ then for every invertible matrix $W$ you have $$p(W^{-1}AW)=W^{-1}p(A)W=0 . $$
Here $p=x^2-x$, you can take $A=\begin{pmatrix}1&0\\0&0\end{pmatrix}$, $W$ any invertible matrix to make a lot of examples.
A: Any projection operator obeys this relation.  It should be intuitive that when you apply this operator again, the projected vector should not change.
One can prove this concretely.  For a unit vector $u$, let $\underline A(a) = a - (u \cdot a)u$. This projects the vector $a$ onto the subspace orthogonal to $u$.  Clearly $\underline A^2(a) = a - (u \cdot a) u - (u \cdot a) u + (u \cdot u)(u \cdot a) u = a - (u \cdot a)u = \underline A(a)$.
