rank of $PQ$, rank of $QP$

Question

Given that $m$ and $n$ are natural numbers and $m < n$. $P$ is an $n{\times}m$ real matrix, and $Q$ is an $m{\times}n$ real matrix. Then which of the following is/are not possible.

a)$\;\text{rank}(PQ)=n$

b)$\;\text{rank}(QP)=m$

c)$\;\text{rank}(PQ)=m$

d)$\;\text{rank}(QP)=[(m+n)/2]$ ,where $[x]$ is defined as the smallest integer greater or equal to $x$.

option a) is not possible because of the theorem $\text{rank}(PQ) \le \min\{\text{rank}(P),\text{rank}(Q)\}$.

Now $[(m+n)/2] > m$, but $QP$ is an $m{\times}m$ matrix, so option d) should not be possible.

The answer is given only option a).

Am I doing any mistake for option d).

Half of the question is being posted here.The middle half is not getting posted.this is half of my written question.I am trying to post the full question. — user514386, Feb 23 '18 at 06:48
https://chat.stackexchange.com/rooms/73566/general-chat , we can chat here. BTW: I've the same doubt for option D. — Learning, Feb 23 '18 at 08:42

quasi · Answer 1 · 2018-02-23T08:09:58.750

It's possible to have $m < n$, but $[(m+n)/2] = m$.

For example, if $m=2$, and $n=3$, then $$[(m+n)/2] = [5/2]=2 = m$$

More generally, if $m,n$ are integers with $m < n$, then $[(m+n)/2] = m$ if and only if $n=m+1$.

As an example to show that options $(b),(c),(d)$ are possible, let $m=2$, let $n=3$, and let $P,Q$ be defined by \begin{align*} P&= \pmatrix{ 1 & 0\\ 0 & 1\\ 0 & 0\\ } \\[8pt] Q&=P^{T} \end{align*} Then $\text{rank}(PQ)=2$, and $\text{rank}(QP)=2$, so only option $(a)$ fails.

Edit:

I misread it.

Usually, the notation $[x]$ is a variant of $\lfloor{x}\rfloor$, so reading quickly, I didn't see that you actually specified ceiling, not floor.

But please check the source. If $[x]$ really is specified as ceiling, then since $m < n$, $$[(m+n)/2] \ge [m+(m+1))/2] = [(2m+1)/2] = [m + (1/2)] = m+1 > m$$ so as you argued, option $(d)$ is not possible.

For the matrices $P,Q$ given above, options $(b)$ and $(c)$ are both satisfied, hence, assuming $[x]$ was truly specified as $\lceil{x}\rceil$, it follows that options $(a)$ and $(d)$ are the only options that must fail.

Hi Quasi , Here [x] is defined to be the smallest integer larger than or equal to x. — user514386, Feb 23 '18 at 07:45

score 0 · Answer 2 · answered Feb 23 '18 at 09:21

Assuming that $Q$ is an $n \times m$ so that $PQ$ and $QP$ both make sense, then remember that $$\text{rank}(QP) = \text{rank}(P) - \dim(\ker (Q) \cap \text{range}(P))$$

Further to this, one can write
$$\text{rank}(PQ) =\text{rank}(Q) - \dim(\ker(P) \cap \text{range}(Q))$$

with no specific reason for equality here.

If we are given that $n$ and the ranks of $P$ and $Q$, $\ker(P)$ and $\text{range}(Q)$ can be any subspaces of ${\mathbb R}^n$ of dimensions $n - \text{rank}(P)$ and $\text{rank}(Q)$ respectively, and their intersection can have any dimension from $\max(\text{rank}(Q) - \text{rank(P)},0)$ to $\min(n - \text{rank}(P),\text{rank}(Q))$.

It is not completely out of the question that these could equate, but WLOG they will not be.

rank of $PQ$, rank of $QP$

2 Answers2