Linear Algebra, Theory And Applications, 2012a

116 ROW OPERATIONS
Proof: Since ker (A) is a subspace, there exists a basis for ker (A) , {x 1 , ··· , x k } . Also
let {Ay 1 , ··· ,Ay l } be a basis for A (F n ). Let u ∈ F n . Then there exist unique scalars c i
such that
l∑
Au = c i Ay i
It follows that
(
A u−
i=1
)
l∑
c i y i = 0
and so the vector in parenthesis is in ker (A). Thus there exist unique b j such that
u =
i=1
l∑
c i y i +
i=1
k∑
b j x j
Since u was arbitrary, this shows {x 1 , ··· , x k , y 1 , ··· , y l } spans F n . If these vectors are
independent, then they will form a basis and the claimed equation will be obtained. Suppose
then that
l∑
k∑
c i y i + b j x j = 0
Apply A to both sides. This yields
i=1
j=1
j=1
l∑
c i Ay i = 0
and so each c i =0. Then the independence of the x j imply each b j =0.
i=1
4.4 Rank And Existence Of Solutions To Linear Systems
Consider the linear system of equations,
Ax = b (4.4)
where A is an m × n matrix, x is a n × 1 column vector, and b is an m × 1 column vector.
Suppose
A = ( )
a 1 ··· a n
where the a k denote the columns of A. Thenx =(x 1 , ··· ,x n ) T is a solution of the system
(4.4), if and only if
x 1 a 1 + ···+ x n a n = b
which says that b is a vector in span (a 1 , ··· , a n ) . This shows that there exists a solution
to the system, (4.4) if and only if b is contained in span (a 1 , ··· , a n ) . In words, there is a
solution to (4.4) if and only if b is in the column space of A. In terms of rank, the following
proposition describes the situation.
Proposition 4.4.1 Let A be an m × n matrix and let b be an m × 1 column vector. Then
there exists a solution to (4.4) if and only if
rank ( A | b ) =rank(A) . (4.5)