Prove that A is idempotent without using Jordan form Suppose $A$ is a $n\times n$ matrix with coefficients in a field , and $\operatorname{rank}(A) + \operatorname{rank}(I_n-A) = n$. How to prove $A^2 = A$ without using Jordan form?
What I know :
For any $v$ we have $v = Av + (I_n - A )v$.
Hence $$V = \operatorname{Im}(A) + \operatorname{Im}(I_n - A)$$
Note that
\begin{align}
n &= \operatorname{dim}(V) = \operatorname{dim}(\operatorname{Im}(A) + \operatorname{Im}(I_n - A)) \\
&= \operatorname{dim}(\operatorname{Im}(A)) +\operatorname{dim}(\operatorname{Im}(I_n - A)) - \operatorname{dim}(\operatorname{Im}(A)\cap\operatorname{Im}(I_n - A))\\
& = n - \operatorname{dim}(\operatorname{Im}(A)\cap\operatorname{Im}(I_n - A)) 
\end{align}
So we have
$V = \operatorname{Im}(A) \oplus \operatorname{Im}(I_n - A)$.
 A: There are also some other facts you can observe about the kernals. As $(I-A)v=0$ implies that $Av=v$, we see that
$$\ker(I-A)\subseteq Im(A)$$
and likewise
$$\ker(A)\subseteq Im(I-A).$$
This implies that $\ker(I-A)\cap \ker(A)\subseteq Im(A)\cap Im(I-A)=0$, so that by the rank-nullity theorem
$$dim(\ker(I-A))+dim(\ker(A))=2n-dim(Im(I-A))-dim(Im(A))=n.$$
Thus $$V=\ker(I-A)\oplus \ker(A)\subseteq Im(A)\oplus Im(I-A)=V,$$
so that $$\ker(I-A)=Im(A),\;\;\;\;\ker(A)=Im(I-A).$$
The first of this implies that $(I-A)A=0$ so that $A-A^2=0$and $A=A^2$ as you desire.
A: First note that in general we have $\ker A \subseteq \operatorname{im}(I-A)$. Indeed, if $Ax = 0$, then $$x = (I-A)x \in \operatorname{im}(I-A).$$
Now it follows
\begin{align}
A^2 = A &\iff A(I-A) = 0 \\
&\iff \operatorname{im}(I-A) \subseteq \ker(A) \\
&\iff \operatorname{im}(I-A) =\ker(A) \\
&\iff \dim \operatorname{im}(I-A) = \dim\ker(A)\\
&\iff \operatorname{rank}(I-A) = n-\operatorname{rank}(A)
\end{align}
where in the last equivalence we used the rank-nullity theorem.
A: if you want to do this in 3 lines
$\operatorname{rank}(A) + \operatorname{rank}(I_n-A) = n$
$\implies \operatorname{dim}\ker A + \operatorname{dim}\ker(I_n-A) = n$ (negate, add $2n$ to each side and apply rank-nullity)
$\implies A$ is diagonalizable with all eigenvalues $\in \big\{0,1\big\}$ since eigenvectors corresponding to disintinct eigenvalues are linearly independent, and the above says they form a basis for $V$.  Thus $A^2=A$.
