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$\mathbf {The \ Problem \ is}:$ 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 .

Git Gud
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Rabi Kumar Chakraborty
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  • A quicker proof of the equality you attained: $$ r(A-B) = r([A - AB] + [AB - B]) \leq r(A - AB) + r(AB - B). $$ – Ben Grossmann Jan 19 '20 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. – Ben Grossmann Jan 19 '20 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 $\ker(I - A) = \operatorname{im}(A)$. – Ben Grossmann Jan 19 '20 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. – Rabi Kumar Chakraborty Jan 19 '20 at 21:21

2 Answers2

2

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

Ben Grossmann
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  • Can you explain "By the rank-nullity theorem...". By applying the rank-nullity theorem to which operators exactly? – Bart Michels Jan 19 '20 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 – Ben Grossmann Jan 19 '20 at 16:20
  • @punctureddusk fixed my answer. Unfortunately, now it's not as short as I would have hoped. – Ben Grossmann Jan 19 '20 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. – Bart Michels Jan 19 '20 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 $U$. – Ben Grossmann Jan 19 '20 at 16:55
  • But the orthogonal complement of $\ker(A)$ w.r.t. the Euclidean inner product is not necessarily $\operatorname{im}(A^T)$, right? – Bart Michels Jan 19 '20 at 17:04
  • @punctureddusk actually, it is! See this post for instance – Ben Grossmann Jan 19 '20 at 17:34
  • @punctureddusk actually, in the paragraph about the intersection of the images I use the fact that $B$ is idempotent. I was very terse, so I'm sure it isn't immediately clear. – Ben Grossmann Jan 19 '20 at 17:47
  • @punctureddusk thanks for the bounty! – Ben Grossmann Jan 24 '20 at 07:36
2

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.

user1551
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