Exercise 3.2.48

Answers

(a): By matrix multiplication, each column of $AB$ is $A$ times the corresponding column of $B$ . So if column $j$ of $B$ is a combination of earlier columns, then column $j$ of $AB$ is the same combination of earlier columns of $AB$ . Then rank $(AB) \leq rank (B)$ . No new pivot columns!
(b): The rank of $B$ is $r = 1$ . Multiplying by $A$ cannot increase this rank. The rank of $AB$ stays the same for $A_{1} = I$ and $B = [\begin{matrix} 1 & 1 \\ 1 & 1 \end{matrix}] .$ It drops to zero for $A_{2} = [\begin{matrix} 1 & 1 \\ - 1 & - 1 \end{matrix}]$ .

mathchakra

2022-01-23 14:01