Principles of Mathematics in Operations Research

128 9 Number Systems

Proposition 9.4.1 Let z,w G C. Then,

(a) z ^ 0 => \z\ > 0 and |0| = 0.

W 1*1 = N-

(c,) |^u)| = |z||i«|.

(dj |ite(z)| ^ |z|, |Im(*)| < \z\.

(e) \z + w\ < \z\ + \w\, [Triangle inequality].

Proof. The first three is trivial. Then,

(d) Let z = a + ib \Im(z)\ = \b\ = y/P < Va^2 + b^2 = \z\.

(e) \z + w\^2 = (z + w)(z + w) = \z\^2 + zw + zw + \w\^2 < (\z\ + \w\)^2

~zw + zw = 2Re(zw) < \2Re(zw)\ •< 2\zw\ => \z + w\ •< \z\ + \w\.

Take positive square roots of both sides, i.e. if a >z 0, b >z 0 and a^2 <

b^2 =4- a •< b. If not, b < a =$> b^2 ^ ab, ba < a^2 =>• b^2 ^ a^2. Contradiction!

•

Theorem 9.4.2 (Schwartz Inequality) Let aj,bj £ C, j = l,...,n.

Then,

J2

a

i

b

i

i=i

-<

Proof. ByO.HB = 0 then bj = 0 V? => LHS = 0; therefore, 0^0.

Assume B y 0 =>

n n

0 r< ]T \Baj - Cbjf = Yl(Bai - Cbj)(BN ~ Cb~)

j=i i=i

= Y/

B2

\

a

i\

2

-Jl

BCa

i

b

i-Y,

CBb

M

+ y

£\

c

\

2

\

b

i\

3 = 1 3 = 1 3 = 1 3 = 1

= B^2 A - B\C\^2 - CBC+ \C\^2 B = B(AB - |C|^2 ).

Thus, .45 >: |C|^2 , since B ^ 0. D

9.5 Euclidean Space

Definition 9.5.1 Let k £ N, we define Rk as the set of all ordered It-
tuples x = (x\,... ,Xk) of real numbers x\,...,Xk- We define {x + y) =
(xi + 2/ii • • • )#*; + Vk)- If a £ R, ax — (ax\,... ,axk). This way Rk be-
comes a vector space over K.
We define an inner product inRk byx-y = 5Zi=1 £J2/J. AndVx £ Rk, x-x X 0.
We define the norm of x £Rk by \x\ = \Jx • x = y^rl=1 x^2 ^.