Is $A^2 = A$ diagonalizable? I found the eigenvalues as $0$ and $1$. The eigenspaces are then the $NS(A)$ and the $CS(A)$, how do I prove that these spaces provide enough information to chose the correct eigenvectors and form a diagonalizable $A$ ?
 A: For any $x \in V$, we have $x = Ax + (x - Ax)$.  Now, note that $A (Ax) = Ax$ so $Ax$ is in the eigenspace for $\lambda = 1$; and $A (x - Ax) = 0$, so $x - Ax$ is in the eigenspace for $\lambda = 0$.  This shows that the eigenvectors of $A$ span $V$.
A: Let $m(X)$ be the minimal polynomial of $A$ and $P(X)=X^2-X$. Then, since $P(A)=0$ we have $m(X) \mid X^2-X$.
This shows that the minimal polynomial of $A$ has distinct roots, and hence $A$ is diagonalisable.
A: You have $A^2 = A.$
If $x$ is in the column space of $A$ then $Ax=x.$ That is because $x = Ay$ for some vector $y,$ so $Ax= A(Ay) = A^2 y = Ay = x.$
If $x$ is not in the column space of $A$ then $x-Ax$ is in the null space of $A.$ That is because $A(x-Ax) = Ax-A^2x = Ax-Ax = 0.$
So every vector $x$ can be written as a member of the column space plus a member of the null space, thus: $x = \underbrace{\quad Ax\quad} {}+{} \underbrace{(x-Ax)}.$
This implies every vector can be written as a sum of an eigenvector with eigenvalue $1$ and an eigenvector with eigenvalue $0.$ In other words, the domain is the direct sum of those two eigenspaces. That means there is a basis of the domain consisting only of eigenvectors. With respect to that basis, the matrix of this linear transformation is diagonal. And every diagonal entry is $0$ or $1.$
