Let \(P,Q\) be \(n\times n\) matrices satisfying \[ P^2=P, \qquad Q^2=Q, \] and suppose \[ I-(P+Q) \] is invertible. Prove that \(P\) and \(Q\) have the same rank.
Solution
Put \[ A\coloneqq I-(P+Q). \]
1. The matrices \(A,P,Q\) satisfy\[ AP=QA. \]
Using \(P^2=P\) gives \[ AP =(I-P-Q)P =P-P^2-QP =-QP. \] Using \(Q^2=Q\) gives \[ QA =Q(I-P-Q) =Q-QP-Q^2 =-QP. \] Hence \(AP=QA\).
2. The matrices \(P\) and \(Q\) are similar:\[ Q=APA^{-1}. \]
By hypothesis, \(A\) is invertible. Multiplying the identity in step 1 on the right by \(A^{-1}\) gives the displayed equality.
3. Therefore\[ \boxed{\operatorname{rank}P=\operatorname{rank}Q}. \]
By step 2, \[ Q=APA^{-1}. \] Left and right multiplication by invertible matrices preserve rank, so \(P\) and \(Q\) have the same rank.
4. Q.E.D.
Step 3 is the required conclusion.