Linear Algebra Solutions 38 - a 11 x 1 ·· a r 1 x r = a...

where A = [ a
ij
] is non–singular. Then the above system of equations can
be solved for X
1
, . . . , X
r
in terms of Y
1
, . . . , Y
r
. Consequently by Theorem
3.2.1
h Y
1
, . . . , Y
r
i = h X
1
, . . . , X
r
i = S.
It follows from problem 11 that Y
1
, . . . , Y
r
is a basis for S .
(ii) We show that all bases for S are given by equations 2. So suppose
that Y
1
, . . . , Y
r
forms a basis for S . Then because X
1
, . . . , X
r
form a basis
for S , we can express Y
1
, . . . , Y
r
in terms of X
1
, . . . , X
r
as in 2, for some
matrix A = [ a
ij
]. We show A is non–singular by demonstrating that the
linear independence of Y
1
, . . . , Y
r
implies that the rows of A are linearly
independent.
So assume
x
1
[ a
11
, . . . , a
1 r
] + · · · + x
r
[ a
r 1
, . . . , a
rr
] = [0 , . . . ,