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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?

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The expression $\mathbf{A} = \left[\vec{A_1} \cdots \vec{A_i} \cdots \vec{A_n}\right]^{\huge{T}}_{m \times n}$ needs to be edited, as it states that $A$ is the transpose of a row vector. –  littleO Sep 2 '13 at 5:16
    
@littleO: Thanks; $\require{enclose} \enclose{horizontalstrike}{\vec{A_n}}$ should be $\vec{A_m}$. I wrote $\mathbf{A}$ with transpose to shuffle out of (gratuitous) elongation and white space. Is this not true: $\mathbf{A} = \left[\vec{A_1} \cdots \vec{A_i} \cdots \vec{A_m}\right]^{\huge{T}}_{m \times n} = \left[\begin{matrix} \vec{A_1} \\ \vdots \\ \vec{A_i} \\ \vdots \\ \vec{A_m} \end{matrix}\right]_{m \times n} ?$ –  LePressentiment Sep 2 '13 at 8:51
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Actually $\left[\vec{A_1} \cdots \vec{A_i} \cdots \vec{A_m}\right]^{\huge{T}} = \left[\begin{matrix} \vec{A_1}^T \\ \vdots \\ \vec{A_i}^T \\ \vdots \\ \vec{A_m}^T \end{matrix}\right]$. Note that, if each $\vec{A_i}$ is $1 \times n$, then $\left[\vec{A_1} \cdots \vec{A_i} \cdots \vec{A_m}\right]$ is a row vector of size $1 \times mn$. –  littleO Sep 2 '13 at 9:03
    
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2 Answers

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 $.

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Thank you very much. +1. I wish I could've upvoted more. Could you please reveal how and why "left multiplication is independent of columns of $B$ and hence $A$ is actually a row transformation"? Also, how can I intuit or naturalise this? At present, I'm enduring/"capitulating" to this via the definition, but I'd like to discern it instinctively. –  LePressentiment Dec 11 '13 at 13:45
    
@LePressentiment "the ith column of AB is obtained from left multiplicating A to the ith column of B", which is irrelvent to column $j\neq i$. That's what I mean "independent" –  Shuchang Dec 11 '13 at 13:58
    
I'd be grateful if you would please answer question $2$ also? –  LePressentiment Mar 10 at 6:17
    
@LePressentiment If $A$ is written as a row vector, then you cannot multiply $B$ anymore. –  Shuchang Mar 10 at 13:28
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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).

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I think there is no abuse of notation, and the last two matrices are exactly the same. You're using block notation. –  littleO Sep 2 '13 at 5:12
    
@littleO One is a 1x2 matrix, while the other is a 2x2 matrix. –  A.P. Sep 2 '13 at 5:14
    
That's not how I understand block notation. If $a_1,a_2$ are column vectors (let's say they're $N \times 1$), then I think $\begin{bmatrix} a_1 & a_2 \end{bmatrix}$ is just a concise notation for the $N \times 2$ matrix whose first column is $a_1$ and whose second column is $a_2$. In any case, I suppose the issue is minor. –  littleO Sep 2 '13 at 5:20
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