Rank of $I_m - X_{m \times m}$ given rank of $X$

Question

I have a matrix $X_{m\times m}$ which is idempotent and has $rank(X) = n < m$. I have for some time now been trying to calculate $rank(I_m - X)$ but have been unable to do so. I should be able to get an exact value for the rank, not just an inequality.

I have established or know the following things.

1. By the rank-nullity theorem $rank(I-A) + dim$ $\mathcal{N}(I - A) = m$.

2. $\forall v \in \mathcal{N}(X)$ we have $(I-X)v = Iv - Xv = Iv = v$.

3. $\forall v \in \mathcal{N}(I - X)$ we have $(I-X)v = 0 \implies Iv = v = Xv$. Together with point 2 this implies that $\mathcal{N}(I-X) \cap \mathcal{N}(X) = {0}$.

4. $dim$ $\mathcal{N}(A) = m - n$.

Any help would be greatly appreciated. Thanks.

Answer 1

If $X$ is idempotent then $(X - I)(X) = 0$ so it can only have eigenvalues $0$ and $1$ and also it must be semisimple (diagonalizable). So you can diagonalize it, say with $a$ 0's and $b$ 1's. The rank of $X$ is $b$. Then the rank of $X - I$ is $a$. So the answer is $$rank(X) + rank(X - I) = m$$

Answer 2

If $X$ is an imdepotent of rank $n$ then $I_m - X$ is a idempotent of rank $m-n$.

For the rank: chech that we have a direct sum decomposition $$\mathbb{R}^m =X(\mathbb{R}^m) \oplus ( I_m - X)(\mathbb{R}^m)$$