TITLE: Proving $\text{rank}(A+B) \leq \text{rank}(A) + \text{rank}(B) - \text{rank}(AB)$ when $A,B$ commute
QUESTION [7 upvotes]: Let $A,B$ be $n \times n$ matrices and $AB=BA$. Show that $$\text{rank}(A+B) \leq \text{rank}(A) + \text{rank}(B) - \text{rank}(AB)$$

Attempt at the solution: 
I first proved that $$R(A+B) \subseteq R(A) + R(B)$$ where R(A) is Range(A). 
Then I used the dimension formula that states that $$\dim(U+V) = \dim(U) + \dim(V) - \dim(U \cap V)$$ so that I get $$r(A+B) \leq r(A) +r(B) - \dim(R(A)\cap R(B))$$
The last part would be to show that $$\dim(R(A)\cap R(B)) = \dim(R(AB))$$ using the fact that $AB=BA$. 
I need help with this last part. Also please critique me if the above solution is acceptable or if there is something wrong with it. 
Thank you!!

REPLY [7 votes]: You do not need $\dim(R(A)\cap R(B)) = \dim(R(AB))$. All you need is $\dim(R(A)\cap R(B)) \ge \dim(R(AB))$, which is easy to prove using $AB=BA$.