Relationship between eigenvector and null space in the projection matrix

Question

If $A^2=A$, then $x-Ax$ is in its eigenspace when $\lambda=0$. But I find it difficult to be sure that $x-Ax$ actually generates an entire eigenspace $\ker A$. The domain of a projection matrix can be written as $\ker A\oplus\operatorname{im} A =\operatorname{dom} A$. But can I explain it with only eigenvectors without using this fact? For example, when $x-Ax\neq a$ and $Aa=0$ for all $x$ then how do I induce a contradiction?

Answer 1

Let us suppose that $A^2=A$. Then $x-Ax\in Ker(A)$ since $A(x-Ax)=Ax-A^2x=(A-A^2)x=0$. Therefore $Im(Id-A)\subseteq ker(A)$. However, any vector $x\in ker(A)$ can be written as $x=x-0=x-Ax\in Im(Id-A)$. Therefore $\ker(A)=Im (Id-A).$