Each of A and B is a real nxn matrix with:
- A+B = I (identity) and:
- rank(A)+rank(B) = n
Does each of those identities always hold?
- A2 = A
- B2 = B
- AB = BA = 0
If so, prove it.
Otherwise, provide a counterexample.
Sylvester's Rank Inequality states that if matrix A is m by n and matrix B is n by p, then rank(A)+rank(B) <= rank(A*B) + n.
We are given rank(A)+rank(B) = n, therefore rank(A*B) = 0. The only matricies with rank 0 are the zero matricies, therefore Identity 3, AB=BA=0, is true.
Then to prove the other two identities, we can create some equalities for A^2 - B^2 and A^2 + B^2:
A^2 - B^2 = (A+B) * (A-B) = I * (A-B) = A - B
A^2 - B^2 = A - B
A+B = I = I^2 = (A+B)^2 = A^2 + AB + BA + B^2 = A^2 + B^2
A^2 + B^2 = A + B
Add the two simplified identities:
(A^2 - B^2) + (A^2 + B^2) = (A - B) + (A + B)
2A^2 = 2A
A^2 = A
This proves Identity 1. Identity 2 is proved similarly with B^2 - A^2 = B - A and A^2 + B^2 = A + B.