Let $M$ be a $7×6$ real matrix. The entries of $M$ in the positions $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$ are changed
to obtain another $7×6$ real matrix $\widetilde{M}$. Suppose that the rank of $\widetilde{M}$ is 4. What could be the rank of $M$?
List all possibilities.
Solution:
First, we will derive a connection between the rank of $M$ and the rank of $\widetilde{M}$. We have,
$$rank(M) = rank(M-\widetilde{M}+\widetilde{M}) \le rank(M-\widetilde{M})+rank(M)$$ Similarly
$$rank(\widetilde M) = rank(\widetilde{M}-M+M) \le rank(\widetilde{M}-M)+rank(M)$$
This shows that $rank(\widetilde{M}) - rank(\widetilde{M}-M) \le rank(M) \le rank(M-\widetilde{M})+rank(\widetilde{M}) $
The matrices $M - \widetilde{M}$ and $\widetilde{M}-M$ can have non-zero entries only in the columns $3$ and $4$ and hence they are of rank atmost $2$. Also, it is given that the rank of $\widetilde{M}$ is equal to $4$. Hence we have $$2 \le rank(M) \le 6.$$
The matrices $M - \widetilde{M}$ and $\widetilde{M}-M$ can have non-zero entries only in the columns $3$ and $4$ and hence they are of rank atmost $2$. Also, it is given that the rank of $\widetilde{M}$ is equal to $4$. Hence we have $$2 \le rank(M) \le 6.$$
Using $0,1$ matrics one can check that all these possibilities are occurring indeed.