Meaning of the matrix product行列積ぎょうれつせき

1Introduction

The product of matrices is not entrywise multiplication. It is defined so that matrices represent composition of linear maps線型写像せんけいしゃぞう in coordinates.

The rule for matrix multiplication can look unnatural at first, but it appears naturally once we require compatibility with the action on column vectors and with composition of maps.

2Terms and definitions

Let $A$A be an $m\times n$m\times n matrix行列ぎょうれつ and $B$B an $n\times p$n\times p matrix. Then $AB$AB is an $m\times p$m\times p matrix with entries

The inner sizes must match.

This formula is not merely a rule to memorize. If $b^{\left(j\right)}$b^{(j)} is the $j$j-th column列れつ of $B$B, then the $j$j-th column of $AB$AB is

Thus $B$B first sends the $j$j-th standard-basis direction to $b^{\left(j\right)}$b^{(j)}, and then $A$A acts on that result. Reading the $i$i-th component gives

3Strategy

First calculate matrix products as tables of numbers, then interpret them as actions on column vectors列れつベクトル. Finally, check that matrix multiplication represents composition of linear maps線型写像せんけいしゃぞう.

4Intuitive explanation

Let the columns of $A$A be $a_1,\dots ,a_n$a_1,\ldots,a_n. For $x=(x_1,\dots ,x_n{)}^T$x=(x_1,\ldots,x_n)^T,

A matrix therefore uses the components of the input as coefficients係数けいすう and forms a linear combination線型結合せんけいけつごう of its columns.

Applying this device twice produces matrix multiplication. Thus $ABx$ABx means: apply $B$B first, then apply $A$A.

5Precise explanation

Regard $B:K^p\to K^n$B:K^p\to K^n and $A:K^n\to K^m$A:K^n\to K^m. For $x\in K^p$x\in K^p,

Therefore the composite map is $A\circ B$A\circ B, and its representation matrix表現行列ひょうげんぎょうれつ is $AB$AB. Since matrices act from the right on vectors, the matrix applied first appears on the right.

6Column interpretation

If the columns of $A$A are $a_1,\dots ,a_n$a_1,\ldots,a_n and $x=(x_1,\dots ,x_n{)}^T$x=(x_1,\ldots,x_n)^T, then

So a matrix uses the components of the input as coefficients and forms a linear combination線型結合せんけいけつごう of its columns. If $b^{\left(j\right)}$b^{(j)} is the $j$j-th column of $B$B, then the $j$j-th column of $AB$AB is

7Composition interpretation

Regard $B:K^p\to K^n$B:K^p\to K^n and $A:K^n\to K^m$A:K^n\to K^m. Then

so the composite map is represented by $AB$AB. In $ABx$ABx, $B$B acts first and $A$A acts second.

8Worked example

Let

Then

Thus $AB$AB and $BA$BA are generally different because the order of composition is different.

9Theorem and proof: matrix product行列積ぎょうれつせき represents composition

Let $A$A be $m\times n$m\times n and $B$B be $n\times p$n\times p. Write the columns of $B$B as $b_1,\dots ,b_p$b_1,\ldots,b_p. For $x\in K^p$x\in K^p,

Applying $A$A and using linearity gives

The columns of $AB$AB are exactly $A{b}_1,\dots ,A{b}_p$Ab_1,\ldots,Ab_p, so $\left(AB\right)x=A\left(Bx\right)$(AB)x=A(Bx).

10Common misunderstandings

• matrix product行列積ぎょうれつせき is not the product of corresponding entries.
• Size compatibility must be checked before calculating.
• $AB=BA$AB=BA is not expected unless there is a special reason.

11Scope

Matrix multiplication is defined so that it agrees with composition of linear maps線型写像せんけいしゃぞう. There are other operations that multiply corresponding entries componentwise, but those are different from the standard matrix product.

12Final forms

13Theorem and proof: matrix product is composition of linear maps

Let $A$A be an $m\times n$m\times n matrix and $B$B an $n\times p$n\times p matrix. Then $AB$AB represents the linear map that applies $B$B first and then $A$A.

To prove this, take $x\in F^p$x\in\mathbb F^p. If the columns of $B$B are $b_1,\dots ,b_p$b_1,\ldots,b_p, then

Applying $A$A and using linearity線型性せんけいせい gives

On the other hand, the $j$j-th column of $AB$AB is $A{b}_j$Ab_j, so $\left(AB\right)x$(AB)x has the same expression. Hence

This proof explains why the order of matrix multiplication reflects the order of composition.

14Exercise links

15Related links