Matrix Multiplication - Product of [Row or Column Vector] and Matrix [Lay P94, Strang P59] From P59 of Intro to Lin Alg, 4th Ed by Strang & P94-95 of Linear Algebra and its Apps by Lay

For relief, I denote all row vectors with superscripts and column with subscripts.  Define $\mathbf{A} = \left[\begin{matrix} 
\vec{a^1} \\
\vdots \\
\vec{a^i} \\
\vdots \\
\vec{a^m}
\end{matrix}\right]_{m \times n}
\& \quad 
\mathbf{B} =
\left[\vec{b_1} \cdots \vec{b_i} \cdots \vec{b_p}\right]_{n \times p}$ ,
  where each of the $m$ $\vec{a^i}$s has size $(1 \times n)$ and each of the $p$ $\vec{b_i}$s size $(n \times 1)$. Then:
$\mathbf{AB} = \mathbf{A}\left[\vec{b_1} \cdots \vec{b_i} \cdots \vec{b_p}\right]_{n \times p} = \left[\mathbf{A}\vec{b_1} \cdots \mathbf{A}\vec{b_i} \cdots \mathbf{A}\vec{b_p}\right]_{m \times p}. \tag{Row}$
$\text{ Also, } \qquad \mathbf{AB} = \left[\begin{matrix} 
\vec{a^1} \\
\vdots \\
\vec{a^i} \\
\vdots \\
\vec{a^m}
\end{matrix}\right]\mathbf{B} 
= \left[\begin{matrix} 
\vec{a^1}\mathbf{B}  \\
\vdots \\
\vec{a^i}\mathbf{B}  \\
\vdots \\
\vec{a^m}\mathbf{B} 
\end{matrix}\right]_{m \times p}. \tag{Coln}$ 

$1.$ In (Row), how and why can $\mathbf{A}$ left-multiply into the column vector form of $\mathbf{B}$ ?
In (Coln), how and why can $\mathbf{B}$ right-multiply the row vector form of $\mathbf{A}$ ?
$2.$ Would someone please explain how $\mathbf{A}$ can be rewritten as a row vector? 
 A: These things are all matrices, so it's standard matrix multiplication. For example,
$$\left[
        \begin{matrix}
        1 & 2\\
        3 & 4\\
        \end{matrix}\right]\cdot \left[
        \begin{matrix}
        1 & 2\\
3 & 4\\
        \end{matrix}\right] = \left[
        \begin{matrix}
        \left[
        \begin{matrix}
        1 & 2\\
        3 & 4\\
        \end{matrix}\right]\cdot\left[
        \begin{matrix}
        1\\
        3\\
        \end{matrix}\right] & \left[
        \begin{matrix}
        1 & 2\\
        3 & 4\\
        \end{matrix}\right]\cdot\left[
        \begin{matrix}
        2\\
        4\\
        \end{matrix}\right]\\
        \end{matrix}\right] = 
\left[
        \begin{matrix}
        \left[
        \begin{matrix}
        7\\
        15\\
        \end{matrix}\right] & \left[
        \begin{matrix}
        10\\
        22\\
        \end{matrix}\right]\\
        \end{matrix}\right]
=\left[
        \begin{matrix}
        7 & 10\\
        15 & 22\\
        \end{matrix}\right].
$$
(Of course, here, there is a slight notational abuse; the last two matrices aren't exactly the same).
A: The verification is trivial. But to give an intuitive explanation, remember this:
Left multiplication of $B$ by matrix $A$ is equivalent to perform a row transformation on $B$. 
Thus we have to guarantee $B$ contains enough rows to be transformed. Consider matrix is actually multilinear, to arrive at your equation $(Row)$.
With the similar reason, right-multiplication of $B$ by matrix $B$ is equivalent to performing a column transformation on  $A$. Thus we must guarantee $A$ contains enough columns to be transformed. This effects your equation $(Column)$.

To see this, take a set of orthonormal basis $e_i,\ldots,e_n$. Remark the following facts.


*

*$Be_i=\begin{pmatrix}b(1, i)\\\vdots\\b_(n,i)\end{pmatrix}$ selects the $i$th column of matrix $B$

*$e_j^TA=\begin{bmatrix} A(j,1) & \cdots & A(j,n) \\
    \end{bmatrix}$ selects the $j$th row of matrix $A$


Next write $\text{ $i$th column of AB } = (AB)e_i=A(Be_i)=Ab_i$.
This means $i$th column of $AB$ is obtained from left-multiplying $A$ by the $i$th column of $B$.
So all the other columns ($\neq i$) of $B$ and $AB$ are unaffected.
Or we could say the left multiplication (by $\mathbf{A}$) is independent of the other columns of $B$.
 Hence $A$ is actually a row transformation. 
Similarly $\text{ $j$th row of AB } = e_j^TAB = (e_j^TA)B = \begin{bmatrix} A(j,1) & \cdots & A(j,n) \\
    \end{bmatrix}B = (\mathbf{A^j})^TB $.
