Question-10
Let \(A\) and \(B\) be two matrices such that \(AB\) is defined. Select the most appropriate statement.
Matrix multiplication doesn’t increase the rank. To see why note that 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 subspace 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: \[ \text{rank}(AB) \leqslant \min(\text{rank}(A), \text{rank}(B)) \]
Thus if we start with a matrix \(A\) and multiply it by some matrix \(B\), the rank of the resulting matrix cannot be more than the rank of \(A\).