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P. 105
Unit 6: Computation Concerning Subspaces
so that if is a linear combination of the , we have Notes
j
r
= b ki i
i 1
r m
= b ki P ij j
i 1 j 1
m r
= b P ij j
ki
j 1 i 1
r
and thus x = b P
j ki ij
i 1
is one possible choice for the x (there may be many).
j
The question of whether = (b ,...,b ) is a linear combination of the , and if so, what the scalars
1 n i
x are, can also be looked at by asking whether the system of equations
i
m
A x i b j , j 1,...,n
ij
i 1
has a solution and what the solutions are. The coefficient matrix of this system of equations is
then n × m matrix B with column vectors ,..., . In unit 5, we discussed the use of elementary
1 m
row operations in solving a system of equations BX = Y. Let us consider one example in which
we adopt both points of view in answering questions about subspaces of F .
n
6.2 Illustrative Examples
4
Example 1: Let us pose the following problem. Let W be the subspace of R spanned by
the vectors
= (1, 2, 2, 1)
1
= (0, 2, 0, 1)
2
= (–2, 0, –4, 3)
3
(a) Prove that , , form a basis for W, i.e., that these vectors are linearly independent.
1 2 3
(b) Let = (b , b , b , b ) be a vector in W. What are the coordinates of relative to the ordered
1 2 3 4
basis { , , }?
1 2 3
(c) Let
' = (1, 0, 2, 0)
1
' = (0, 2, 0, 1)
2
' = (0, 0, 0, 3)
3
Show that ' , ' , ' form a basis for W.
1 2 3
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