Why are matrix and scalar multiplication denoted the same way and treated as the same operation in standard mathematical notation? This is always a source of confusion for me because they have completely different properties (specifically commutativity). Multiplying a 1x1 matrix by an NxN matrix isn't even generally equivalent to multiplying an NxN matrix by a scalar. (The former is not even always defined.) Wouldn't it be clearer to consider these to be completely unrelated operations and use completely different notation to represent them?

  • $\begingroup$ They are not "treated as the same operation"; we call them both "multiplication", but that's not the same as "treat[ing them] as the same operation". We call addition of vectors "sum", just like sum of numbers, even when it has nothing to do with sums of numbers (the set of positive reals is a vector space over the reals with vector addition $\mathbf{u}\oplus\mathbf{v}=uv$ and scalar product $\alpha\odot\mathbf{v}=v^{\alpha}$). We call them "multiplication" because they have enough similar properties to warrant it. $\endgroup$ – Arturo Magidin Mar 29 '11 at 20:12
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    $\begingroup$ That said, "multiplication" of matrices is best thought of as and analogue of "composition" (just like composition of functions). It's just that the set of $N\times N$ matrices, for a fixed $N$, form a ring en.wikipedia.org/wiki/Ring_%28mathematics%29 under addition of matrices and matrix composition/multiplication, and the second operation of rings is always called "multiplication". $\endgroup$ – Arturo Magidin Mar 29 '11 at 20:17
  • $\begingroup$ @Arturo: Good point, but vector addition maps to scalar addition in a much more transparent way (it's just the component-wise sum) and preserves important properties like commutativity. It's subjective, but I think scalar vs. matrix multiplication stretches things much further than scalar vs. vector addition. Thus from a notational clarity point of view, denoting vector and scalar addition the same way is more reasonable than denoting scalar and matrix multiplication the same way. $\endgroup$ – dsimcha Mar 29 '11 at 20:17
  • $\begingroup$ No, vector addition is not always "component-wise sum". That's the point of my example above. For what it may be worth, "scalar multiplication" is definitely different from "multiplication". Multiplication is a binary operation on a set (takes two arguments from the set, returns an element of the set), whereas "scalar multiplication" is not an operation under that definition. In General Algebra, "scalar multiplication" is viewed as an entire family of unary operations instead. $\endgroup$ – Arturo Magidin Mar 29 '11 at 20:20
  • $\begingroup$ By "scalar multiplication" I understood the vector space map that takes a scalar and a vector and returns a vector; if you meant "multiplication of 'scalars'/'numbers'", then my last comment probably reads like nonsense. $\endgroup$ – Arturo Magidin Mar 29 '11 at 20:35

The product of matrices is defined so that it corresponds to the composition of the corresponding linear maps. One can derive the usual formula for matrix multiplication from this fact alone. This should be covered in every good linear algebra textbook, e.g. Axler's Linear Algebra Done Right. $\:$ See also Arturo Magidin's answer here. So your question reduces to why composition of maps is denoted the same as multiplication. One answer is that rings arise naturally as subrings of linear maps on their underlying additive groups (left regular representation). This is a ring-theoretic analog of the Cayley represention of a group as subgroups of permutation, by acting on itself by left multiplication. This allows us to view "functions" as "numbers" and exploit operator theoretic techniques such as factoring characteristic polynomials and differential and difference operators (recurrences), etc. The point of the common notation is to emphasize this common ring structure so that one may exploit it by reusing similar techniques where they apply.

Examples of such techniques abound. For some examples of operator algebra see here, here, here. See also here, here where the fibonacci recurrence is recast into linear system form, yielding an addition formula and fast computation algorithm by repeating squaring of the shift matrix.

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    $\begingroup$ I'm sure this answer makes sense to a theoretical mathematician, but I asked the question from an engineer/applied mathematician's perspective. To me "multiplication", when unqualified, means scalar multiplication. From my perspective (this is admittedly subjective) if something is called multiplication, its properties should be similar to scalar multiplication in ways that are obvious in everyday contexts and don't require a Ph.D. in pure mathematics to appreciate. $\endgroup$ – dsimcha Mar 29 '11 at 20:28
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    $\begingroup$ @dsimcha: It's always a good idea to say something about your level of knowledge when asking a question on a general level math site. Do you know any abstract algebra, e.g. do you know the definition of a ring? This is undergrad math, not Ph.D. $\endgroup$ – Bill Dubuque Mar 29 '11 at 20:31
  • $\begingroup$ No, I only know math related to engineering and probability/statistics. I know virtually nothing about the more theoretical branches. I've heard of abstract algebra, but know absolutely nothing about what it is. In hindsight I should have clarified this in my original post. Then again, your post makes me believe the reason matrix multiplication is called "multiplication" is much deeper than scientists, engineers, statisticians and generally people who aren't full-fledged mathematicians have the background to appreciate. $\endgroup$ – dsimcha Mar 29 '11 at 20:46
  • $\begingroup$ @dsimcha: The basic ideas are accessible to engineers. I've added some links to prior questions which give some examples. $\endgroup$ – Bill Dubuque Mar 29 '11 at 20:57
  • $\begingroup$ @dsimcha: would you agree that, forgetting all that phd math stuff, scalar multiplication and vector multiplication are similar to each other than either are to, say, vector addition? $\endgroup$ – Mitch Mar 29 '11 at 22:25

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