Step 1: Understanding the Topic
This question relates the properties of two matrices, $P$ and $Q$, given that their product is the zero matrix. It involves concepts of rank, null space, and column space, and how they are constrained by the matrix dimensions and the product condition.
Step 2: Key Approach - Rank-Nullity and Subspace Relations
The condition $PQ=0$ implies that for any vector $x \in \mathbb{R}^6$, the vector $Qx$ is in the null space of $P$. This means the column space of $Q$ is a subspace of the null space of $P$. We can use this relationship along with the Rank-Nullity Theorem to derive an inequality involving the ranks of $P$ and $Q$.
Step 3: Detailed Derivation and Analysis
A. Deriving the Rank Inequality:
The condition $PQ=0_{6 \times 6}$ implies that every vector in the column space of $Q$ is mapped to the zero vector by $P$. Therefore, the column space of $Q$ is contained within the null space of $P$:
\[ \text{ColumnSpace}(Q) \subseteq \text{NullSpace}(P) \]
This subspace relationship implies an inequality between their dimensions:
\[ \dim(\text{ColumnSpace}(Q)) \le \dim(\text{NullSpace}(P)) \]
By definition, $\text{rank}(Q) = \dim(\text{ColumnSpace}(Q))$ and $\text{nullity}(P) = \dim(\text{NullSpace}(P))$. So:
\[ \text{rank}(Q) \le \text{nullity}(P) \]
Now we apply the Rank-Nullity Theorem to matrix $P$, which is a $6 \times 4$ matrix (it has 4 columns):
\[ \text{rank}(P) + \text{nullity}(P) = (\text{number of columns of } P) = 4 \]
From this, we can express nullity$(P)$ as: $\text{nullity}(P) = 4 - \text{rank}(P)$.
Substituting this into our inequality:
\[ \text{rank}(Q) \le 4 - \text{rank}(P) \]
Rearranging this gives Sylvester's Rank Inequality for this case:
\[ \text{rank}(P) + \text{rank}(Q) \le 4 \]
B. Evaluating the Options:
The derived, mathematically correct inequality is $\text{rank}(P) + \text{rank}(Q) \le 4$. Let's examine the options:
(A) and (B) describe other subspace relationships which are not necessarily true.
(D) is a specific case ($r(P) + r(Q) = 4$) which can happen but is not guaranteed. For example, if $P=0$ and $Q=0$, then $r(P)+r(Q)=0 \le 4$.
(C) states $r(P) + r(Q) \ge 4$. This contradicts our derived result $r(P) + r(Q) \le 4$, unless equality holds.
There seems to be an error in the question's options or the provided answer key, as the correct conclusion is a 'less than or equal to' inequality, while the provided answer is a 'greater than or equal to' inequality.
Step 4: Final Answer
Based on standard linear algebra theorems, the correct relationship derived from the premises is $\text{rank}(P) + \text{rank}(Q) \le 4$. None of the provided options accurately reflect this general result. The option (C) provided in the key is inconsistent with this derivation.