Principles of Mathematics in Operations Research

216 Solutions

t

/,

wis

,r w ^

0'

•**>*

A

.ttT*

R(A)=Span(yi.y 2 )

Fig. S.4. The range and null spaces of A — [j/i |y2J2/3]

where Vjv

-1

-e, S, =

-10 0

: '•• 0

_1 ... _i

0"

0_

) &II [l,...,l] = e

T.

Thus, 11(A) = Span {j/i, • • • , yn-i}- A/-(A) = Span {£}, where

* =

r-vwi

L

J

i J

i—i

i

Moreover,

1l(AT) = Span {-yn, -yi, ••• , -yn-2} •

And finally, Af(AT) = Span {5//} = Span {[1, • • • , 1]T} = Span {e}. We

have Af(A) = (ftM)-^1 - since A/"U) = -^(A- ) (therefore, 1Z(AT) = 11(A)

by the Fundamental Theorem of Linear Algebra-part 2) in this particular

exercise.