What do we gain or lose, conceptually, if we consider scalar multiplication as a special form of matrix multiplication?


The question bothers me since I have been reading about dilations and scaling of geometrical objects in Paul Lockhart's book "Measurement". Geometrically, dilation is a transformation that stretches an object in one dimension by a certain factor. Analogously, the linear transformation $$ \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & \lambda \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} $$ "stretches" the third component by the factor $\lambda$. Scaling is a geometric transformation that stretches an object in all dimensions by a certain factor. Analogously, the linear transformation $$ \begin{pmatrix} \lambda & 0 & 0 \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{pmatrix} \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} $$ "stretches" all three components by the factor $\lambda$. This, however, can be written more succinctly using scalar multiplication: $$ \lambda \cdot \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}. $$ In fact, every scalar multiplication can be expressed as a multiplication with a special matrix, and it turns out to be a mere shortcut. On the face of it, this observation is not very spectacular; however, it raises interesting philosophical and conceptual questions as to the foundations of linear algebra.

For example, if scalar multiplication is only a nice-to-have shortcut, then isn't it in fact superfluous conceptually? Currently, scalar multiplication is taught as if it was a distinct concept, independent of matrix multiplication. What would change if we got rid of this shortcut? What could alternative axioms of vector spaces and moduls look like? What about linear transformations? What is easier, what is harder$-$not to write down, but conceptually?

I know that this topic is very broad, but I would like to collect opinions, ideas, examples.

  • 1
    $\begingroup$ The definition of a linear transformation refers to scalar multiplication. And matrices are often introduced as a concise way to describe linear transformations. I think that's a nice way to organize these ideas. To me it seems backwards to introduce matrix multiplication before introducing linear transformations. $\endgroup$ – littleO Apr 16 '16 at 9:29
  • $\begingroup$ Scalar multiplication can "move around" in matrix multiplications: $kABC = k(ABC) = (kA) BC = A(kB)C = AB(kC)$ $\endgroup$ – peterwhy Apr 16 '16 at 9:30
  • $\begingroup$ When the vector space has an uncountable basis, vectors or linear transformations in general are not representable by matrices. So, you simply cannot treat scalar multiplication as a matrix operation. $\endgroup$ – user1551 Apr 16 '16 at 11:48
  • $\begingroup$ No, it's not true that scalars are superfluous. They are integral to the definition of the vector space. That is, choosing a different field, will result in a different vector space, even though the underlying Abelian group will be the same. For example, the Abelian group $(\mathbb R, +)$ forms a one-dimensional vector space over the field $\mathbb R$, but an infinite dimensional vector space over $\mathbb Q$. $\endgroup$ – M. Vinay Apr 22 '16 at 14:16
  • $\begingroup$ Similarly, a field extension is a vector space over all its subfields. So if you have $F_1 \subset F_2 \subset F_3$, with $F_1$ proper in $F_2$, and $F_2$ proper in $F_3$, then $F_3$ forms a vector space over $F_2$ as well as over $F_1$, and these two spaces are non-isomorphic, thought they are isomorphic as Abelian groups. It's the scalars that make all the difference. $\endgroup$ – M. Vinay Apr 22 '16 at 14:17

I disagree that "scalar multiplication is only a nice-to-have shortcut", or that is "superfluous conceptually". In fact the very definition of a vector space $V$ requires there to be a scalar multiplication.

After that comes the concept of a linear transformation, which again requires the scalar multiplication to be defined. Matrix multiplication is a convenient way to represent these, and that too only in case of finite dimensional vector spaces (or certain infinite dimensional vector spaces).

So defining matrix multiplication and then saying that scalar multiplication is a special case is putting the cart in front of the horse in my opinion.

  • $\begingroup$ Am I right that you disagree, because you don't like the order of introduction of concepts, rather than because it will not work conceptually? In fact, as you say, you could define matrix multiplication first. And when you have it, you can introduce the axioms of vector spaces. Both ways are equivalent. $\endgroup$ – Björn Friedrich Apr 16 '16 at 10:44
  • $\begingroup$ What do you define the multiplication of a matrix with? To define matrix multiplication first, you'll need to define vectors as columns of scalars, whereas the abstract definition of a vector space encompasses so much more. $\endgroup$ – Seven Apr 16 '16 at 11:30
  • $\begingroup$ To define multiplication of a matrix with a "vector" only requires that the vector comes from an abelian group. The scalar multiplication is not needed. $\endgroup$ – Björn Friedrich Apr 16 '16 at 11:54
  • 1
    $\begingroup$ @BjörnFriedrich I don't get your last comment. How are you defining a vector? And with this definition, how are you defining matrix multiplication? $\endgroup$ – Seven Apr 16 '16 at 12:02

Note: Some discussion with Siddharth has revealed a flaw in my earlier answer, which I've corrected now (with a complete reversal of conclusion).

The set of all endomorphisms of an Abelian group $(V, +)$ forms a ring $\operatorname{End}(G)$ with pointwise addition as the ring addition and composition of mappings as the ring multiplication. If there exists a subring $L \subseteq \operatorname{End}(G)$ with center $F = \operatorname{Z}(L)$, such that $F$ is a field and $F \cap \operatorname{End}(G) = (1)$, the ideal generated by the identity endomorphism, then $V$ is a vector space over the field $F$ with $L$ as its ring of linear operators.

Why does this work? [Edit: I'm not sure the following is entirely correct. I'll have to work on it further and update the answer]. In one direction (vector space $\to$ above characterization), it's obvious that the linear operators of a vector space $V$ do form a subring of the endomorphism ring of the Abelian group $(V, +)$, and that the subset of "scalar" operators form a subfield of this ring. As shown here (beautifully), the scalar operators are exactly the central elements of the ring of linear operators. Conversely, it's not hard to show that the Abelian group $(V, +)$ forms a (traditional) vector space over the field $F$ defined above, in a manner that makes $L$ (isomorphic to) the ring of all linear operators of this space.

  • 2
    $\begingroup$ This is an interesting way to look at vector spaces. And in some sense, it does look at scalar multiplication as a special case of matrix multiplication (specifically, if we replace matrices by $L$ and scalars by $Z(L)$). $\endgroup$ – Seven Apr 16 '16 at 23:56
  • $\begingroup$ @Seven There was a problem in the earlier answer. I'm in the process of correcting it. But it's obvious that you cannot avoid scalars in the definition. Different fields will turn the same Abelian group into different, non-isomorphic vector spaces. $\endgroup$ – M. Vinay Apr 22 '16 at 14:13
  • 1
    $\begingroup$ I don't seem to see any flaw in the arguments presented here. Can you elaborate? What exactly do you mean by "different fields will turn the same Abelian group into different, non-isomorphic vector spaces"? Can you give an example to show what you mean? $\endgroup$ – Seven Apr 22 '16 at 14:57
  • $\begingroup$ @Seven Earlier, I had hastily (and wrongly) concluded that all endomorphisms of the group $(V, +)$ would be endomorphisms of "the" vector space formed $(V, +)$ over a field. Turns out I was wrong, and $(V, +)$ can form several different vector spaces (over different fields), and these will all have (in general) different endomorphism rings. But in all cases, these endomorphisms will be subrings of the full endomorphism ring of the group. $\endgroup$ – M. Vinay Apr 22 '16 at 15:01
  • $\begingroup$ Yes I looked at them. Taking the centers of subrings of the full endomorphism ring seems a good idea. Is there a difference between doing that or taking subfields of $Z(\text{End}(G))$? $\endgroup$ – Seven Apr 22 '16 at 15:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.