$\mathrm{rank}(A)+\mathrm{rank}(I-A)=n$ for $A$ idempotent matrix

Question

Let $A$ be a square matrix of order $n$. Prove that if $A^2=A$ then $\mathrm{rank}(A)+\mathrm{rank}(I-A)=n$.

I tried to bring the $A$ over to the left hand side and factorise it out, but do not know how to proceed. please help.

Answer 1

Hint. Show that under the given condition, the following holds:

$$\ker A = \operatorname{im} (I -A) $$

Answer 2

Let rank $(A) = r$ and $\lambda$ be an eigenvalue of $A$. Since $A$ is idempotent then \begin{align*} \lambda x = A x = A^2 x = A.Ax = A\lambda x = \lambda Ax = \lambda^2 x \end{align*} meaning that $\lambda =0$ or $1$. So the eigenvalues of $A$ are either $0$ or $1$. Writing the eigenvalue decomposition of $A$ as \begin{align*} A = U^T \Lambda U \end{align*} where $U$ is the orthogonal matrix of eigenvectors of $A$ and $\Lambda$ is the diagonal matrix of eigenvalues of $A$, we get \begin{align*} I- A &= I - U^T \Lambda U\\ &= U^TU - U^T \Lambda U\\ &= U^T(I-\Lambda)U. \end{align*} Note that since rank $(A) = r$ then only $r$ of the elements of the diagonal of $\Lambda$ are $1$ and the rest are zero. This implies that only $n-r$ elements of $I-\Lambda$ are $1$ and the rest are zero. So, rank $(I-A) = n-r$ and you have the desired result.