Linear Programming Lecture Notes - Penn State Personal Web Server

which clearly has a solution for all a, b, and c. Another way of seeing this is to note that the
matrix:
(3.40)
⎡
1
A = ⎣ 1
1
0
⎤
0
1 ⎦
0 1 1
is invertible.
The following theorem on the size of a basis in R n is outside the scope of this course. A
proof can be found in [Lan87].
Theorem 3.37. If X is a basis of R n , then X contains precisely n vectors.
Exercise 34. Show that the vectors
⎡
x1 = ⎣ 1
⎤ ⎡
2⎦
, x2 = ⎣
3
4
⎤ ⎡
5⎦
, x3 = ⎣
6
7
⎤
8⎦
9
are not a basis for R 3 . [Hint: See exercise 33.]
We will use the following lemma, which is related to the notion of the basis of R n when
we come to our formal method for solving linear programming problems.
Lemma 3.38. Let {x1, . . . , xm+1} be a linearly dependent set of vectors in R n and let
X = {x1, . . . , xm} be a linearly independent set. Further assume that xm+1 = 0. Assume
α1, . . . , αm+1 are a set of scalars, not all zero, so that
(3.41)
m+1
i=1
αixi = 0
For any j ∈ {1, . . . , m} such that αj = 0, if we replace xj in the set X with xm+1, then this
new set of vectors is linearly independent.
Proof. Clearly αm+1 cannot be zero, since we assumed that X is linearly independent.
Since xm+1 = 0, we know there is at least one other αi (i = 1, . . . , m) not zero. Without
loss of generality, assume that αm = 0 (if not, rearrange the vectors to make this true).
We can solve for xm+1 using this equation to obtain:
(3.42) xm+1 =
m
i=1
− αi
xi
αm+1
Suppose, without loss of generality, we replace xm by xm+1 in X . We now proceed by
contradiction. Assume this new set is linearly dependent. There there exists constants
β1, . . . , βm−1, βm+1, not all zero, such that:
(3.43) β1x1 + · · · + βm−1xm−1 + βm+1xm+1 = 0.
Again, we know that βm+1 = 0 since the set {x1, . . . , xm−1} is linearly independent because
X is linearly independent. Then using Equation 3.42 we see that:
m
(3.44) β1x1 + · · · + βm−1xm−1 + βm+1
= 0.
i=1
− αi
xi
αm+1
42