Page 137 - A Course in Linear Algebra with Applications

5.1: Existence of a Basis 121

First notice that these two vectors are linearly indepen-
2
dent and generate R , so that they form a basis. We need to
find scalars c and d such that

1)-«(!)-(!

This amounts to solving the linear system

c + 3d = 2
c + Ad = 3

i
The unique solution s c = — 1, d = 1, and hence the coordi-
-1
nate vector is ,

Coordinate vectors provide us with a method of testing a
subset of an arbitrary finitely generated vector space for linear
dependence.

Theorem 5.1.8
Let {vi,..., v n } be an ordered basis of a vector space V. Let
Ui,..., u m be a set of vectors in V whose coordinate vectors
with respect to the given ordered basis are Xi,..., X m respec-
tively. Then {u±,... ,u m } is linearly dependent if and only
if the number of pivots of the matrix A = [X1IX2I... \X m] is
less than m.
Proof
Write Uj = ^ " = 1 ajiVj\ then the entries of Xi are an,..., a ni,
so the (j,i) entry of A is a^. If ci,... c m are any scalars,
,
then

m n n m
v
c x ui H h c mu m = ^T Ci (^2 ajiVj) = ^2(^2 a ji°i) j-
i=l j = l j=l i = l