TITLE: Matrix multiplication: interpreting and understanding the process
QUESTION [53 upvotes]: I have just watched the first half of the 3rd lecture of Gilbert Strang on the open course ware with link:
http://ocw.mit.edu/courses/mathematics/18-06-linear-algebra-spring-2010/video-lectures/
It seems that with a matrix multiplication $AB=C$, that the entries as scalars, are formed from the dot product computations of the rows of $A$ with the columns of $B$. Visual interpretations from mechanics of overlpaing forces come to mind immediately because that is the source for the dot product (inner product).
I see the rows of $C$ as being the dot product of the rows of $B$, with the dot product of a particular row of $A$. Similar to the above and it is easy to see this from the individual entries in the matrix $C$ as to which elements change to give which dot products.
For understanding matrix multiplication there is the geometrical interpretation, that the matrix multiplication is a change in the reference system since matrix $B$ can be seen as a transormation operator for rotation, scalling, reflection and skew. It is easy to see this by constructing example $B$ matrices with these effects on $A$. This decomposition is a strong argument and is strongly convincing of its generality. This interpreation is strong but not smooth because I would find smoother an explanation which would be an interpretation begining from the dot product of vectors and using this to explain the process and the interpretation of the results (one which is a bit easier to see without many examples of the putting numbers in and seeing what comes out which students go through).
I can hope that sticking to dot products throughout the explanation and THEN seeing how these can be seen to produce scalings, rotations, and skewings would be better. But, after some simple graphical examples I saw this doesn't work as the order of the columns in matrix $B$ are important and don't show in the graphical representation. 
The best explanation I can find is at Yahoo Answers. It is convincing but a bit disappointing (explains why this approach preserves the "composition of linear transformations"; thanks @Arturo Magidin). So the question is: Why does matrix multiplication happen as it does, and are there good practical examples to support it? Preferably not via rotations/scalings/skews (thanks @lhf).

REPLY [14 votes]: I think part of the problem people have with getting used to linear transformations vs. matrices is that they have probably never seen an example of a linear transformation defined without reference to a matrix or a basis. So here is such an example. Let $V$ be the vector space of real polynomials of degree at most $3$, and let $f : V \to V$ be the derivative.
$V$ does not come equipped with a natural choice of basis. You might argue that $\{ 1, x, x^2, x^3 \}$ is natural, but it's only convenient: there's no reason to privilege this basis over $\{ 1, (x+c), (x+c)^2, (x+c)^3 \}$ for any $c \in \mathbb{R}$ (and, depending on what my definitions are, it is literally impossible to do so). More generally, $\{ a_0(x), a_1(x), a_2(x), a_3(x) \}$ is a basis for any collection of polynomials $a_i$ of degree $i$.
$V$ also does not come equipped with a natural choice of dot product, so there's no way to include those in the discussion without making an arbitrary choice. It really is just a vector space equipped with a linear transformation.
Since we want to talk about composition, let's write down a second linear transformation. $g : V \to V$ will send a polynomial $p(x)$ to the polynomial $p(x + 1)$. Note that, once again, I do not need to refer to a basis to define $g$. 
Then the abstract composition $gf : V \to V$ is well-defined; it sends a polynomial $p(x)$ to the polynomial $p'(x + 1)$. I don't need to refer to a basis or multiply any matrices to see this; all I am doing is composing two functions.
Now let's do everything in a particular basis to see that we get the same answer using the correct and natural definition of matrix multiplication. We'll use the basis $ \{ 1, x, x^2, x^3 \}$. In this basis $f$ has matrix
$$\left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\\
 0 & 0 & 2 & 0 \\\
 0 & 0 & 0 & 3 \\\
 0 & 0 & 0 & 0 \end{array} \right]$$
and $g$ has matrix
$$\left[ \begin{array}{cccc} 1 & 1 & 1 & 1 \\\
 0 & 1 & 2 & 3 \\\
 0 & 0 & 1 & 3 \\\
 0 & 0 & 0 & 1 \end{array} \right].$$
Now I encourage you to go through all the generalities in Arturo's post in this example to verify that $gf$ has the matrix it is supposed to have.