TITLE: Prove that if $\mathrm{rank}(A-I_n)=p$ and $\mathrm{rank}(B-I_n)=q$, then $\mathrm{rank}(AB-I_n)\leq p+q$
QUESTION [0 upvotes]: Let $A,B\in M_n\mathbb{(R)}$ be matrices with $\mathrm{rank}(A-I_n)=p$ and $\mathrm{rank}(B-I_n)=q$. Show that $\mathrm{rank}(AB-I_n)\leq p+q$.
I have no idea how to relate the two rank given to solve the question stated.

REPLY [1 votes]: We have $\DeclareMathOperator{\rank}{rank}$
$$
\rank(AB - I) =\\ 
\rank(AB - A + A - I)=\\
\rank(A(B-I) + A-I) \leq\\
\min\{\rank(A),\rank(B-I)\} + \rank(A-I) \leq \\
\rank(B-I) + \rank(A-I) =\\
p+q
$$
{"set_name": "stack_exchange", "score": 0, "question_id": 1055405}