Projection Matrixes, $A,B$ such that $\text{Id} - (A+B)$ is invertible

Question

This question was on an old qual exam and I have been stuck on it:

Let $A,B$ be two real $5x5$ matrices such that $A^2=A , B^2 = B$ and $\text{Id} - (A+B)$ is invertible. Show Rank($A$)=Rank($B$).

My biggest question is why is $A,B$ being $5x5$ needed? I don't see why this wouldn't generalize to any square matrix with the above properties.

I have tried to use that $\mathbb{R}^5 = \text{Im}(A)\bigoplus\text{Ker}(A)$ and similarly for $B$ and tried to show nullity(A) = nullity(B) to no avail. I believe the strongest assumption given is that $\text{Id} - (A+B)$ is invertible, but I am unsure how to use this to prove Rank($A$) = Rank($B$).

Answer 1

The $5\times 5$ assumption is indeed irrelevant. As a hint, try writing $Id-(A+B)$ as $(Id-A)-B$.

A full proof is hidden below.

Let's say our matrices are $n\times n$, and $A$ has rank $a$ and $B$ has rank $b$. Note then that $Id-A$ is a projection of rank $n-a$, and so $(Id-A)-B$ has rank at most $n-a+b$ (its image is contained in the sum of the image of $Id-A$ and the image of $B$). Since $(Id-A)-B$ is invertible, we have $n\leq n-a+b$ and so $a\leq b$. But swapping the roles of $A$ and $B$, we also get $b\leq a$, and so $a=b$.

Answer 2

You don't need $5\times 5$, $n\times n$ works.

You also don't need $\Bbb R$, any field will work.

As you say, consider $\Bbb R^n=\text{im}\,A+\ker A$. On $\text{im}\, A$, $I-A$ vanishes, so for $I-A-B$ to be invertible, $B$ must be injective on $\text{im}\, A$. Therefore $\text{rank}\,B\ge\dim \text{im}\,A=\text{rank}\,A$.