Question

Let \(m, n \in \mathbb{N}\) such that \(m<n\) and \(P_{m \times n}(\mathbb{R})\) and \(Q_{n \times m}(\mathbb{R})\) are matrices over real numbers and let \(\rho(V)\) denotes the rank of the matrix V. Then, which of the following are NOT possible.
A. \( \rho(PQ) = n \)
B. \( \rho(QP) = m \)
C. \( \rho(PQ) = m \)
D. \( \rho(QP) = \lfloor(m+n)/2\rfloor \), where \(\lfloor \rfloor\) is the greatest integer function

• A. A and D only
• B. B and C only
• C. A, C and D only
• D. A, B and C only

Hint:

Remember this fundamental rule: the rank of a product of matrices, \(AB\), can never be greater than the rank of either \(A\) or \(B\). This simple rule is often sufficient to quickly eliminate impossible scenarios in rank-related problems.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question deals with the rank of the product of two non-square matrices. We will use the fundamental properties of matrix rank, specifically how it relates to matrix dimensions and products. 

Step 2: Key Formula or Approach: 
A key property of the rank of a product of two matrices \( A \) and \( B \) is:

\[ \rho(AB) \le \min(\rho(A), \rho(B)) \]

Also, the rank of a matrix cannot exceed the number of its rows or columns. 
For \(P_{m \times n}\), \( \rho(P) \le \min(m, n) = m \) (since \(m < n\)). 
For \(Q_{n \times m}\), \( \rho(Q) \le \min(n, m) = m \). 

Step 3: Detailed Explanation: 
Let's analyze each statement: 
A. \( \rho(PQ) = n \): 
The matrix product \(PQ\) has dimensions \( (m \times n) \times (n \times m) = m \times m \). The maximum possible rank of an \(m \times m\) matrix is \(m\). We are given that \( m < n \). Therefore, it is impossible for the rank of \(PQ\) to be \(n\). 
Also, \( \rho(PQ) \le \min(\rho(P), \rho(Q)) \le \min(m, m) = m \). Since \(m < n\), \( \rho(PQ) \) must be less than \(n\). Thus, \( \rho(PQ) = n \) is NOT possible. 
B. \( \rho(QP) = m \): 
The matrix product \(QP\) has dimensions \( (n \times m) \times (m \times n) = n \times n \). We know \( \rho(QP) \le \min(\rho(Q), \rho(P)) \le m \). So the rank is at most \(m\). It is possible to construct matrices \(P\) and \(Q\) such that \( \rho(P) = \rho(Q) = m \) and \( \rho(QP) = m \). For example, let \(P\) be a matrix with \(m\) linearly independent rows and \(Q\) be its pseudoinverse. This statement is possible. 
C. \( \rho(PQ) = m \): 
The product \(PQ\) is an \(m \times m\) matrix. Its rank can be at most \(m\). It is possible to construct \(P\) and \(Q\) such that \( \rho(P) = \rho(Q) = m \) and \( \rho(PQ) = m \). This statement is possible. 
D. \( \rho(QP) = \lfloor (m+n)/2 \rfloor \): 
We know that \( \rho(QP) \le m \). For this statement to be possible, we must have \( \lfloor (m+n)/2 \rfloor \le m \). However, since \( m < n \), we can choose \(n\) to be large enough to violate this. For example, let \( m = 2 \) and \( n = 5 \). Then \( m < n \). The required rank would be \( \rho(QP) = \lfloor (2+5)/2 \rfloor = \lfloor 3.5 \rfloor = 3 \). But we know that \( \rho(QP) \le m = 2 \). Since \( 3 > 2 \), this is a contradiction. Because we can find a case where this is impossible, the statement is generally NOT possible. 
Step 4: Final Answer: 
Statements A and D describe situations that are not generally possible. Therefore, the correct option is (A).