Page 136 - 35Linear Algebra

136 Matrices

In other words, addition just adds corresponding entries in two matrices,
n
n
and scalar multiplication multiplies every entry. Notice that M = R is just
1
the vector space of column vectors.
Recall that we can multiply an r × k matrix by a k × 1 column vector to
produce a r × 1 column vector using the rule
k
X
i j
MV = m v .
j
j=1
This suggests the rule for multiplying an r × k matrix M by a k × s
matrix N: our k ×s matrix N consists of s column vectors side-by-side, each
of dimension k × 1. We can multiply our r × k matrix M by each of these s
column vectors using the rule we already know, obtaining s column vectors
each of dimension r × 1. If we place these s column vectors side-by-side, we
obtain an r × s matrix MN.
That is, let
 1 1 1 
n 1 n 2 · · · n s
 2 n 2 · · · n 2
n
 1 2 s
N =  . . . . . . 
 . . . 
n k n k · · · n k
2
s
1
and call the columns N 1 through N s :
 1   1   1 
n 1 n 2 n s
 n 2   n 2   n 2 
N 1 =  .  , N 2 =  .  , . . . , N s =  .  .



1 
2 
s 
. . .
 .   .   . 
n k 1 n k 2 n k s
Then
   
| | | | | |
MN = M  N 1 N 2 · · · N s  =  MN 1 MN 2 · · · MN s 
| | | | | |
i
i
Concisely: If M = (m ) for i = 1, . . . , r; j = 1, . . . , k and N = (n ) for
j j
i
i = 1, . . . , k; j = 1, . . . , s, then MN = L where L = (` ) for i = i, . . . , r; j =
j
1, . . . , s is given by
k
X p
i
i
` = m n .
j p j
p=1
136