Homework Set # 3 SOLUTIONS Math 453
Section 6.1
19. Prove that a linear transformation is one-to-one if and only if the null space of T is cfw_0.
Solution:
Suppose T is one-to-one, and suppose there is some x N ull(T ). Then by denition of null
space, T (
MATH 453: Solutions to Homework 9
10.20
Proof. Suppose the original transportation problem is
Minimize
cij xij
i,j
subject to
xij = si
j
xij = dj
i
xij 0
s
d
such that every si , dj and cij are rational numbers. Suppose si = si,1 , dj = dj,1 and
i,2
j,2
c
MATH 453: Solutions to Homework 7
6.19
Proof. Suppose the strong duality theorem is known. We would like to prove the
Farkas Lemma. Consider the following LP and its dual.
Minimize
0x
subject to Ax = b
x 0.
Maximize
yb
subject to yA 0
y unrestricted.
Supp
MATH 453: Solutions to Homework 3
2.33
Proof. We rst show that if f is concave, then its hypograph is convex. Take
two arbitrary points (x1 , y1 ) and (x2 , y2 ) from its hypograph, then by the denition
y1 f (x1 ) and y2 f (x2 ). To show the convexity of
Homework Set # 2 SOLUTIONS Math 453
Section 5.6
6. Suppose that is a norm on Rp and that A is a real nonsingular p p matrix. Dene
x A := Ax and prove that A is a norm.
Solution:
First we need to show that x A = 0 i x = 0, and otherwise x A > 0. Note that
Homework Set # 3 SOLUTIONS Math 453
Section 7.5
11. If w is complex, show that I (2/wH w)(wwH ) is unitary.
Solution: Note that
I (2/wH w)(wwH )
H
I (2/wH w)(wwH ) = I (2/wH w)(wwH ) I (2/wH w)(wwH )
= I (4/wH w)(wwH + (4/(wH w)2 )(wwH wwH )
= I (4/wH w)
MATH 453: Solutions to Homework 5
3.25
Proof. First we rewrite the linear program into the standard form by introducing
slack variables and adding new variables for the unrestricted variables:
Minimize
2x1 2x1 x2 + x2
subject to x1 x1 3x2 + 3x2 x3 = 3
2x1
MATH 453: Solutions to Homework 8
12.1
Proof. Denote by fi,j be the amount of ow from vertex i to j. The maximal
ow is the following: f1,2 = 2, f1,3 = 4, f2,5 = 2, f3,4 = 2, f3,7 = 2, f4,5 = 1,
f4,6 = 1, f6,5 = 1, f5,8 = 4, f7,8 = 2. The size of the maxim
Homework Set # 1 SOLUTIONS Math 453
Section 5.1
12. Part b - u =< 4 2 1 >T and v =< 4 0 3 >T . Find an equation of the form ax+by+cz = d
satised by the entries x, y and z in w =< x y z >T = u + v
solution: x = 4 4, y = 2, and z = + 3 so 3x + 4z = 8 and so