Problem-8

rank

Question

Let \(A\) and \(B\) be two matrices such that \(AB\) is defined. Select the most appropriate statement.

  1. \(\text{rank}(AB) = \text{rank}(A) + \text{rank}(B)\)

  2. \(\text{rank}(AB) = \text{rank}(A) \cdot \text{rank}(B)\)

  3. \(\text{rank}(AB) = \max(\text{rank}(A), \text{rank}(B))\)

  4. \(\text{rank}(AB) = \min(\text{rank}(A), \text{rank}(B))\)

  5. \(\text{rank}(AB) \leqslant \max(\text{rank}(A), \text{rank}(B))\)

  6. \(\text{rank}(AB) \leqslant \min(\text{rank}(A), \text{rank}(B))\)

Solution

Option (f) is correct.

We can view matrix multiplication in the following way. If \(B = \begin{bmatrix}b_1 & \cdots & b_n\end{bmatrix}\) has \(n\) columns, then \(AB\) can be written as: \[ AB = \begin{bmatrix} Ab_1 & \cdots & Ab_n \end{bmatrix} \] We see that the \(i^{th}\) column of \(AB\) is a linear combination of the columns of \(A\), with the multipliers coming from the \(i^{th}\) column of \(b\). Thus the column space of \(AB\) is a subset of the column space of \(A\). It follows that \(\text{rank}(AB) \leqslant \text{rank}(A)\).

We could also view matrix multiplication in another way. If \(A = \begin{bmatrix}r_1^T\\\vdots\\r_m^T\end{bmatrix}\) has \(m\) rows, then \(AB\) can be written as: \[ AB = \begin{bmatrix} r_1^TB\\ \vdots\\ r_m^TB \end{bmatrix} \] We see that the \(i^{th}\) row of \(AB\) is a linear combination of the rows of \(B\), with the multipliers coming form the \(i^{th}\) row of \(A\). Thus the row space of \(AB\) is a subspace of the row space of \(B\). It follows that \(\text{rank}(AB) \leqslant \text{rank}(B)\).

Combining the two inequalities, we have: \[ \text{rank}(AB) \leqslant \min(\text{rank}(A), \text{rank}(B)) \]