Page 65 - Linear Algebra Done Right

The Matrix of a Linear Map
is a basis of U. Consider linear maps S : U → V and T : V → W. The
composition TS is a linear map from U to W. How can M(TS) be
computed from M(T) and M(S)? The nicest solution to this question 51
would be to have the following pretty relationship:
3.11 M(TS) =M(T)M(S).
So far, however, the right side of this equation does not make sense
because we have not yet defined the product of two matrices. We will
choose a definition of matrix multiplication that forces the equation
above to hold. Let’s see how to do this.
Let
   
a 1,1 ... a 1,n b 1,1 ... b 1,p
 . .   . . 
M(T) =  . . . .  and M(S) =  . . . .  .
   
a m,1 ... a m,n b n,1 ... b n,p
For k ∈{1,...,p}, we have
n

TSu k = T( b r,k v r )
r=1
n

= b r,k Tv r
r=1
n m

= b r,k a j,r w j
r=1 j=1
m n

= ( a j,r b r,k )w j .
j=1 r=1
Thus M(TS) is the m-by-p matrix whose entry in row j, column k
n
equals r=1 a j,r b r,k .
Now it’s clear how to define matrix multiplication so that 3.11 holds. You probably learned
Namely, if A is an m-by-n matrix with entries a j,k and B is an n-by-p this definition of matrix
matrix with entries b j,k , then AB is defined to be the m-by-p matrix multiplication in an
whose entry in row j, column k, equals earlier course, although
n you may not have seen

a j,r b r,k . this motivation for it.
r=1
In other words, the entry in row j, column k,of AB is computed by
taking row j of A and column k of B, multiplying together correspond-
ing entries, and then summing. Note that we define the product of two