# A problem regarding the rank of a matrix

$$\mathbf {The \ Problem \ is}:$$ If $$A$$ and $$B$$ be two $$n\times n$$ square matrix such that $$A^2=A$$ and $$B^2=B,$$ then show that $$r(A-B)=r(A-AB)+r(AB-B)$$ where $$r(A)$$ denotes rank of the square matrix $$A.$$

$$\mathbf {My \ approach} :$$ Actually I have tried that, from the two equations, $$A(A-B)=(A-AB)$$ and $$(A-B)B=(AB-B)$$, we have $$r(A-AB)+r(AB-B) \leq r(A)+r(B)$$ and again $$A(AB-B)=0$$ and $$(A-AB)B=0$$, then $$r(A)+r(AB-B)\leq n$$ and $$r(B)+(A-AB)B\leq n$$, but I can't draw any further conclusion to show that $$r(A-AB)+r(AB-B) \leq r(A-B)$$ .

And the other side is obvious by the rank-inequality $$r(P+Q)\leq r(P)+r(Q).$$

A small hint is warmly appreciated .

• A quicker proof of the equality you attained: $$r(A-B) = r([A - AB] + [AB - B]) \leq r(A - AB) + r(AB - B).$$ Jan 19, 2020 at 15:33
• For the other direction, it suffices to show that $(A - AB),(AB - B)$ have trivially-intersecting row-spaces and columns spaces (which leads to a quick proof), but that might be a bit too strong a tool for this problem. Jan 19, 2020 at 15:35
• I suspect that we might be able to come up with an elegant proof starting with $A - AB = (I - A)B$, $AB - B = A(I - B)$ and using the fact that $\ker(I - A) = \operatorname{im}(A)$. Jan 19, 2020 at 17:45
• @Omnomnomnom,Sir,I was also thinking about this part, by using the fact that the vector space $V$ is the direct sum of the eigenspaces of the matrices $A$ and $B$ corresponding to the eigenvalues $0$ and $1$ respectively. But, I couldn't get further. Jan 19, 2020 at 21:21

We want to show that $$r(A-B) = r(A - AB) + r(AB - B).$$

By the claim proven below, it suffices to show that $$A - AB$$, $$AB - B$$ have trivially intersecting row-spaces and trivially intersecting kernels.

To show that the row-spaces intersect trivially, note that $$\operatorname{im}(A - AB)^T = \operatorname{im}((I - B)^TA^T) \subset \operatorname{im}(I-B)^T,\\ \operatorname{im}(AB - B)^T = \operatorname{im}(B^T(I - A^T)) \subset \operatorname{im}(B)^T.\\$$ However, we have $$\operatorname{im}(B)^T \cap \operatorname{im}(I-B)^T = \{0\}$$. To see this: if $$x$$ is in the first space, then $$(I - B^T)x = 0$$ which means that $$B^Tx = x$$. If $$x$$ is in the second space, then $$B^Tx = 0$$. For both to be true, we must have $$x = 0$$.

We now show that the kernels intersect trivially. Suppose that $$x \in \ker(A-B)$$, i.e. $$(A-B)x = 0$$. This can be rewritten as $$(A - B)x = 0 \implies Ax - Bx = 0 \implies Ax = Bx.$$ It follows that $$(A - AB)x = Ax - A(Bx) = Ax - A(Ax) = (Ax - A^2)x = 0.$$ Similarly, $$(AB - B)x = A(Bx) - (Bx) = A(Ax) - (Ax) = (A^2 - A)x = 0.$$ So, $$x \in \ker(A-B)$$ implies that $$x \in \ker(A - AB)$$ and $$x \in \ker(AB - B)$$. The conclusion follows.

Claim: given $$P,Q$$ with $$\operatorname{im}(P^T)\cap \operatorname{im}(Q^T) = \{0\}$$, we have $$r(P + Q) = r(P) + r(Q) \iff \ker(P + Q) = \ker(P) \cap \ker(Q).$$

Proof: $$\ker(P + Q) = \ker(P) \cap \ker(Q) \iff\\ \operatorname{im}(P^T + Q^T) = \operatorname{im}(P^T) + \operatorname{im}(Q^T) \iff\\ \dim \operatorname{im}(P^T + Q^T) = \dim \operatorname{im}(P^T) + \dim \operatorname{im}(Q^T) - \dim[\operatorname{im}(P^T) \cap \operatorname{im}(Q^T)] \iff\\ \dim \operatorname{im}(P^T + Q^T) = \dim \operatorname{im}(P^T) + \dim \operatorname{im}(Q^T) \implies\\ r(P + Q) = r(P) + r(Q).$$

• Can you explain "By the rank-nullity theorem...". By applying the rank-nullity theorem to which operators exactly? Jan 19, 2020 at 16:07
• @punctureddusk to $A-B$. My justification was in order for the rank of $A-B$ to be as large as possible, its kernel must be as small as possible. Now that I reread it though, I see that's too informal. I do think that it is true that $$r(P+Q) = r(P) + r(Q) \iff \ker(P+Q) = \ker(P) + \ker(Q)$$ though the proof eludes me at the moment Jan 19, 2020 at 16:20
• @punctureddusk fixed my answer. Unfortunately, now it's not as short as I would have hoped. Jan 19, 2020 at 16:41
• Can you explain some more the first equivalence in the proof? I'm sorry, this might just me being away from linear algebra for too long. Jan 19, 2020 at 16:53
• @punctureddusk for subspaces $U$ and $V$, $(U \cap V)^\perp = U^\perp + V^\perp$, where $U^\perp$ is the orthogonal complement of $U$. Jan 19, 2020 at 16:55

Let $$F$$ be the underlying field, $$V=F^n,\,A'=I-A$$ and $$B'=I-B$$. Then $$A'$$ and $$B'$$ are also projectors and $$AA'=A'A=BB'=B'B=0$$. We have two observations:

1. $$r(AB')+r(A'B)=r(AB'-A'B)$$:
• Since $$AB'V\cap A'BV\subseteq AV\cap A'V=0$$, we have $$AB'V\cap A'BV=0$$. Hence \begin{aligned} r(AB')+r(A'B)&=\dim(AB'V)+\dim(A'BV)\\ &=\dim(AB'V)+\dim(A'BV)-\dim(AB'V\cap A'BV)\\ &=\dim(AB'V+A'BV). \end{aligned}
• For any $$x,y\in V$$, we have $$AB'x+A'By=(AB'-A'B)(B'x-By)$$. Hence $$AB'V+A'BV\subseteq (AB'-A'B)V$$ and $$\dim(AB'V+A'BV) \le\dim\left((AB'-A'B)V\right) =r(AB'-A'B) \le r(AB')+r(A'B).$$
2. $$r(AB'-A'B)=r(A-B)$$: by rank-nullity thm, it suffices to show that $$\ker(AB'-A'B)=\ker(A-B)$$:
• Suppose $$(AB'-A'B)x=0$$. Left-multiply both sides by $$A$$, we get $$AB'x=0$$. Subtract this equation from the previous one, we obtain $$A'Bx=0$$ too. Now $$AB'x=0$$ and $$A'Bx=0$$ imply that $$Ax=ABx$$ and $$Bx=ABx$$ respectively. Hence $$(A-B)x=0$$.
• Conversely, suppose $$(A-B)x=0$$. Then $$Ax=Bx$$ and $$A'x=B'x$$. Hence $$(AB'-A'B)x=AB'x-A'Bx=AA'x-A'Ax=0$$.

The result now follows from 1 and 2.