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).

Answer 1

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.

Answer 2

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.