Page 37 - Numerical Methods for Chemical Engineering

26 1 Linear algebra

N
Comparing this to the product of an M × N matrix C and w ∈ yields
 N 
   
c 11 c 12 ... c 1N w 1
c 1 j w j

 c 21 c 22 ... c 2N   w 2   j=1 
    . 
Cw=  . . .   .  =  . . 
 . . . . . .   .   
.
 N

c Mj w j
c M1 c M2 ... c MN w N
j=1
N P
 

a 1k b kj w j
 j=1 k=1 . 


=  . .  (1.130)

 N P 

a Mk b kj w j
j=1 k=1
We therefore have the following rule for matrix multiplication to form the M × N matrix
product C = AB of an M ×P matrix A and a P × N matrix B:
P

c ij = a ik b kj (1.131)
k=1
To obtain c ij , we sum the products of the elements of A in row i with those of B in column
j. The matrix product AB is defined only if the number of columns of A equals the number
of rows of B.
We also note that, in general, matrix multiplication is not commutative,
AB = BA (1.132)
From this rule for matrix multiplication, we obtain the following relation for the transpose
of a product of two equal-sized square matrices,
T
T
(AB) = B A T (1.133)
Vector spaces and basis sets
Wemustdiscussonemoretopicbeforeconsideringtheexistenceanduniquenessofsolutions
[2]
[1]
N
to linear systems: the use of basis sets. The set of vectors {b , b ,..., b [P] } in is said
to be linearly independent if there exists no set of real scalars { c 1 , c 2 ,..., c P } except that
of all zeros, c j = 0, j = 1, 2,..., P, such that
c 1 b [1] + c 2 b [2] +· · · + c P b [P] = 0 (1.134)
In other words, if a set is linearly independent, no member of the set can be expressed as
N
[2]
[1]
a linear combination of the others. For vectors in , a set {b , b ,..., b [P] } can be
linearly independent only if P ≤ N.
[1]
[2]
If P = N, we say that the set of N linearly independent vectors {b , b ,..., b [N] }
N
N
forms a basis set for . Then any vector v ∈ may be expressed as a linear combination
v = c 1 b [1] + c 2 b [2] +· · · + c N b [N] c j ∈ (1.135)
A common question is