Dot products in the context of linear algebra and matrix multiplication I've been self-teaching myself linear algebra from Linear Algebra and its Applications 4th from D. Lay. I'm about 8 sections deep and I've had this bothersome feeling regarding the section describing the process of multiplying matrix $A$ and vector $\mathbf x$:

The first entry in the product $A \mathbf x$ is a sum of products (sometimes called a dot product), using the first row of $A$ and the entries in $\mathbf x$. That is,
  $$\begin{bmatrix} 2 & 3 & 4 \\ \phantom{0} \\ \phantom{0} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} 2x_1 + 3x_2 + 4x_3 \\ \phantom{0} \\ \phantom{0} \end{bmatrix}$$

This rolled into some examples:

  
*
  
*$\begin{bmatrix}
  1 & 2 & -1 \\
  0 & -5 & 3
\end{bmatrix}
\begin{bmatrix}
  4 \\ 3 \\ 7
\end{bmatrix}
=
\begin{bmatrix}
  1 \cdot 4 + 2 \cdot 3 + (-1) \cdot 7 \\
  0 \cdot 4 + (-5) \cdot 3 + 3 \cdot 7
\end{bmatrix}
=
\begin{bmatrix}
  3 \\ 6
\end{bmatrix}$
  
*$\begin{bmatrix}
  2 & -3 \\
  8 & 0 \\
  -5 & 2
\end{bmatrix}
\begin{bmatrix}
  4 \\ 7
\end{bmatrix}
=
\begin{bmatrix}
  2 \cdot 4 + (-3) \cdot 7 \\
  8 \cdot 4+ 0 \cdot 7 \\
  (-5) \cdot 4 + 2 \cdot 7
\end{bmatrix}
=
\begin{bmatrix}
  -13 \\ 32 \\ -6
\end{bmatrix}$
  
*$\begin{bmatrix}
  1 & 0 & 0 \\
  0 & 1 & 0 \\
  0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
  r \\ s \\ t
\end{bmatrix}
=
\begin{bmatrix}
  1 \cdot r + 0 \cdot s + 0 \cdot t \\
  0 \cdot r + 1 \cdot s + 0 \cdot t \\
  0 \cdot r + 0 \cdot s + 1 \cdot t \\
\end{bmatrix}
=
\begin{bmatrix}
  r \\ s \\ t
\end{bmatrix}$

I don't know if it's the radical change in syntax or if I'm just plain missing something. The dot products (multivar calc/physics) I'm familiar with result in a scalar answer at the end of the day. As far as I understand, a matrix in itself is a “bundle” of vectors each occupying a column. I get the feeling I'm trying to relate a concept when there isn't a relationship at all I guess. Maybe someone can clarify.
 A: Right - what they're saying is that each (scalar) entry of the matrix can be regarded as a dot product. If $A\vec{x}=\vec{y}$, they are saying that the first entry of $\vec{y}$ is the dot product of the first row of matrix $A$ with the vector $\vec{x}$; the second entry is the dot product of the second row of $A$ with $\vec{x}$; etc.
A: We can relate the dot product to matrix multiplication as follows:
First of all, if you have two column vectors $u=(u_1,u_2,u_3)^T$ and $v=(v_1,v_2,v_3)^T$ ($T$ here means transpose), then the dot product is given by
$$
v\cdot u=
 \begin{bmatrix}
–u^T–
\end{bmatrix}\begin{bmatrix}
|\\
v\\
|
\end{bmatrix}
$$
Conversely, to calculate the entries of the product $AB$ of matrices $A$ and $B$, you could say that the entry in the $i^{th}$ row and $j^{th}$ column is the dot product of the $i^{th}$ row of $A$ with the $j^{th}$ column of $B$.
So for example, take the product
$$
\begin{bmatrix}
1 & 2 & -1\\
0 & -5 & 3\\
\end{bmatrix}
\begin{bmatrix}
4 \\
3 \\
7
\end{bmatrix}
$$
The top entry will be $(1,2,-1)\cdot (4,3,7)$, and the bottom entry will be $(0,-5,3)\cdot (4,3,7)$.
As you said, matrices can be thought of as bundles of vectors.  The tricky bit is wrapping your head around the fact that rows and columns play opposite roles depending on whether the matrix is being multiplied from the right or the left.
A: Basically, multiplying a matrix $A$ by a vector $\vec{x}$ is like taking each row vector stored in $A$ and computing its dot product with the column vector $\vec{x}$, then storing the resulting scalars vertically in a vector $A\vec{x}$. Note that in your first example, $2x_1+3x_2+4x_3$ is a scalar. By slightly abusing the notation, let $\vec{v}_1,\vec{v}_2,...,\vec{v}_m$ be the $m$ row vectors in an $m\times n$ matrix $A$. Then to multiply $A$ by a vector $\vec{x}$ of length $n$, we have:
$$\left[
\begin{array}{c}
\longleftarrow \vec{v}_1 \longrightarrow \\
\longleftarrow \vec{v}_2 \longrightarrow\\
 \vdots \\
\longleftarrow \vec{v}_m \longrightarrow\\
\end{array} 
\right] \vec{x}
=
\left[
\begin{array}{c}
\vec{v}_1 \cdot \vec{x} \\
\vec{v}_2\cdot \vec{x} \\ 
\vdots \\
\vec{v}_m\cdot \vec{x}\\
\end{array}
\right]$$
A: Eventually you learn that the matrix product $AB$ is what must be defined in order that the product of the matrices of a composite is the matrix of the composite: that is: $L_1: \mathbb{R}^n \rightarrow \mathbb{R}^p$ and $L_2: \mathbb{R}^p \rightarrow \mathbb{R}^m$ then the composite $L_2 \circ L_1: \mathbb{R}^n \rightarrow \mathbb{R}^m$ has $[L_2 \circ L_1] = [L_2][L_1]$ where $[L_1]$ is $m \times p$ and $[L_2]$ is $p \times n$ and $[L_2 \circ L_1]$ is $m \times n$. Moreover, if you want a formula $[L]_{ij} = e_i \cdot L(e_j)$. This relation can be used to define the matrix product. Equivalently, the product can be understood to arise from studying substitutions system of linear equations into another. Historically, the subsitution idea came first. But, the logical necessity of the product for the sake of naturalness with composition is a good reason to appreciate the matrix product as it is defined. The way Lay breaks it into several sections is to help you gain computational mastery and to appreciate how $Ax$ is a weighted linear combination of the columns of $A$ by the components of $x$. This is a fantastically useful observation as you continue to study questions of linear independence and spanning. By itself, without a discussion of linear transformations, I will grant you it all seems a bit adhoc.
