I want to find $AB$ where
\begin{gather} A=\begin{bmatrix} 1 & 2 & 0 \\ 0 & 5 & 2 \\ 4& 7 &0 \end{bmatrix} \hspace{1.2em}, \hspace{1.2em} B=\text{span} \{ \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}\} = \text{span} \{b_1,b_2,b_3\} \end{gather}
Do I just multiply $A$ with each column vector of B such that my answer would be $\text{span} \{Ab_1,Ab_2,Ab_3\}$ ?
This might seem obvious, but I need to know whether my method is correct. Thanks in advance!
An element $x\in\mathrm{span}\{b_1,b_2,b_3\}$ is of the form $x=c_1b_1+c_2b_2+c_3b_3$ for some constants $c_1,c_2,c_3.$
By distributivity, $Ax=A(c_1b_1+c_2b_2+c_3b_3)=c_1(Ab_1)+c_2(Ab_2)+c_3(Ab_3)\in\mathrm{span}\{Ab_1,Ab_2,Ab_3\}$.
Similarly, any element of $\mathrm{span}\{Ab_1,Ab_2,Ab_3\}$ can be written as $A$ times an element of $\mathrm{span}\{b_1,b_2,b_3\}$ by using the same coefficients. So you are correct.