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I'm sure everyone already thought about this at least one time. Why matrix multiplication is not defined the way showed below?

$$\left( \begin{array}{ccc} a_{11} & a_{12} & \ldots \\ a_{21} & a_{22} & \ldots \\ \vdots & \vdots & \ddots \end{array} \right) \cdot \left( \begin{array}{ccc} b_{11} & b_{12} & \ldots \\ b_{21} & b_{22} & \ldots \\ \vdots & \vdots & \ddots \end{array} \right) = \left( \begin{array}{ccc} a_{11}\cdot b_{11} & a_{12}\cdot b_{12} & \ldots \\ a_{21}\cdot b_{21} & a_{22}\cdot b_{22} & \ldots \\ \vdots & \vdots & \ddots \end{array} \right) $$

I know this definition has it's limitations. The product only works with same order matrix and any matrix with some zero entry won't be invertible. But this definition is associative, commutative, has identity element, the distributive works, is simpler and more intuitive.

I have no problem about the classic definition, but this definition also has good properties, why it is never used?

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  • $\begingroup$ It is used sometimes. This is the dot product of two vectors, essentially. Sometimes you want to consider this dot product between two matrices. Matrix multiplication as defined usually allows you to compute the composition of linear functions. $\endgroup$
    – user21725
    Apr 25, 2013 at 1:22
  • $\begingroup$ Questions to consider / possible duplicates: here, here, and here. $\endgroup$
    – Zev Chonoles
    Apr 25, 2013 at 1:23
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    $\begingroup$ I don't know if any of these links are duplicates, because they accept the usual product and try to understand it. I'm trying to understand why the definition above is not the usual product. $\endgroup$
    – diff_math
    Apr 25, 2013 at 1:30
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    $\begingroup$ I believe part of the reason is that it wouldn't be very interesting, because matrices would just be long lists of number without any different structure. That's the same reason we don't use componentwise multiplication for vectors: it's just not very useful. $\endgroup$
    – Javier
    Apr 25, 2013 at 1:33
  • $\begingroup$ References and explanations of the definition of matrix multiplication have been given here many times before, e.g. by Bill D and by Arturo M. $\endgroup$
    – Math Gems
    Apr 25, 2013 at 1:38

2 Answers 2

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The matrix multiplication we use is defined that way because it corresponds to the composition of linear maps. Recall that, given a vector space $V$ over $K$ with basis $(e_1,\ldots,e_n)$, and a vector space $W$ over $K$ with basis $(f_1,\ldots,f_m)$, we have a natural isomorphism $\eta:Hom_K(V,W) \rightarrow M_{m\times n}(K)$. The map $\eta$ simply sends a linear map to its matrix representation in terms of the two given bases.

This map is more than an isomorphism of vector spaces: it also preserves the algebra structure, in the sense that composition of linear maps is sent to multiplication of the corresponding matrices.

For example, if you were to compute the effect of the composite operations $\mathbb{R}^3 \xrightarrow{p} \mathbb{R}^2 \xrightarrow{r} \mathbb{R}^2$ in terms of their respective matrices, where $p$ is a projection and $r$ a rotation, you'd simply have to multiply the two matrices together.

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    $\begingroup$ Composition of matrices is just a name, you could it call whatever you want. I just think that the product of matrices is rather artificial. It's not really a product, it's a very specific process that we call product. $\endgroup$
    – diff_math
    Apr 25, 2013 at 1:35
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    $\begingroup$ @diff_math I guess that depends on what you consider worthy of the name "product". $\endgroup$
    – Alex Provost
    Apr 25, 2013 at 1:58
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    $\begingroup$ If we consider the matrix addition and subtraction operations, the natural product(multiplication) would be the one I showed. And the classical product should be called by another name. All the theory with matrices and linear transformations would be exactly the same. $\endgroup$
    – diff_math
    Apr 25, 2013 at 2:07
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    $\begingroup$ @diff_math I guess that depends on your definition of natural :-) $\endgroup$
    – Alex Provost
    Apr 25, 2013 at 2:22
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    $\begingroup$ @diff_math: but then it makes no sense to write the elements as matrices. There's a name for the ring with the product you describe: the direct product of $mn$ copies of the base ring. So we can use that product whenever we want, without adding meaningless row/column structure. $\endgroup$
    – Nick Matteo
    Apr 25, 2013 at 12:18
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Suppose we used your proposed product, along with the usual addition and subtraction of matrices. Then all the algebra of $m\times n$ matrices would be the same as if we just used vectors of length $mn$. (In more detail, you can convert any matrix to a vector by just writing the rows of the matrix, one after the other, as a single long row. And this rewriting process would preserve all the algebraic structure, namely addition, subtraction, and your multiplication.) So this algebra of matrices would not really use the 2-dimensional array structure of matrices at all; it would just be a curious ay of writing component-by-component operations on vectors.

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  • $\begingroup$ I totally agree. That is why Im not excluing the classical product, I just think that "my" definition should be the product and the classical product should be called by any other name. This way, the special structure about matrices would shine with this special operation. $\endgroup$
    – diff_math
    Apr 25, 2013 at 2:35
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    $\begingroup$ Although you won't succeed to redefine standard terminology, you may be pleased to know that "your" product has been used and has a name. It's called the Hadamard product. $\endgroup$
    – Andreas Blass
    Apr 26, 2013 at 21:04

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