Matrix Multiplication - Equality of Row-Column and Column-Row (Strang P71)

Question

Source: P71 of Strang's IoLA, 4th ed, and Wikipedia. How does $(I) = (O)$? I can't conciliate them.

$\bbox[5px,border:2px solid grey]{\text{Inner Product or Row $\cdot$ Column :}}$

$\mathbf{AB} = \left[\begin{matrix} \vec{A_1} \\ \vdots \\ \vec{A_i} \\ \vdots \\ \vec{A_m} \end{matrix}\right]_{m \times n} \left[\vec{B_1} \cdots \vec{B_i} \cdots \vec{B_p}\right]_{n \times p}$ $= \left[\begin{matrix} \vec{A_1}\cdot\vec{B_1} & \cdots & \vec{A_1}\cdot\vec{B_j} & \cdots & \vec{A_1} \cdot \vec{B_p} \\ \vdots & \cdots & \vdots & \cdots & \vdots \\ \vec{A_i}\cdot\vec{B_1} & \cdots & \vec{A_i}\cdot\vec{B_j} & \cdots & \vec{A_i}\cdot \vec{B_p} \\ \vdots & \cdots & \vdots & \cdots & \vdots \\ \vec{A_m}\cdot\vec{B_1} & \cdots & \vec{A_m}\cdot\vec{B_j} & \cdots & \vec{A_m} \cdot\vec{B_p} \\ \end{matrix}\right]$ $\quad \color{Green}{\text{(I)}}$

$\bbox[5px,border:2px solid grey]{\text{Outer Product or Column $\cdot$ Row :}}$

For outer product ($\neq$ inner product), $\cdot$ is defined as the operation of multiplying the $k$th row of A (of size $m \times 1$) with the $kth$ column of B (of size $1 \times p$). This effects a new matrix (of size $m \times p$).

$\mathbf{AB} = \left[\vec{A_1} \cdots \vec{A_k} \cdots \vec{A_n}\right]_{m \times n} \left[\begin{matrix} \vec{B_1} \\ \vdots \\ \vec{B_k} \\ \vdots \\ \vec{B_n} \end{matrix}\right]_{n \times p} := {\left[\require{cancel}\xcancel{\vec{A_1} \cdot \vec{B_1} + \cdots + \vec{A_k} \cdot \vec{B_k} + \cdots + \vec{A_n} \cdot \vec{B_n}} \right]} $
$ = \vec{A_1}\vec{B_1} + \cdots + \vec{A_k}\vec{B_k} + \cdots + \vec{A_n}\vec{B_n} \quad \color{green}{(O)} $

Answer 1

Your inner product rule and (I) is the definition of the matrix product.

For (O), if you write the identity matrix as $I_n = \sum_{i=1}^n e_i e_i^T$ for $(e_1,...,e_n)$ the canonical basis then, because the matrix product is distributive, $$AB = AI_nB = \sum_{i=1}^n (A e_i) (e_i^TB).$$ This is exactly (O) by checking that $A e_i$ is indeed the $i$-th row of $A$ and $e_i^TB$ indeed the $i$-th column of $B$.

Answer 2

EDIT: This response was given for an older version of the question. In the current notation, $\odot$ is the same as multiplying a column vector by a row vector to get a matrix.

In order to see why the two matrices are the same, you'll need to go one level deeper on the definitions of both $\cdot$ and $\odot$ so that everything is written out in terms of the entries of the two original matrices, along with normal real-number addition and multiplication.

What you'll find, roughly speaking, is that the sums hidden in the definition of the $\cdot$ product will be provided explicitly in the $\odot$ formulation. More explicitly, the $k^\text{th}$ $\odot$ product in the summation can be thought of as showing the $k^\text{th}$ term of the $\cdot$ products "all at once".