Page 214 - 35Linear Algebra

214 Basis and Dimension

Theorem 11.0.1. Let S = {v 1 , . . . , v n } be a basis for a vector space V .
Then every vector w ∈ V can be written uniquely as a linear combination of
vectors in the basis S:
1
n
w = c v 1 + · · · + c v n .
Proof. Since S is a basis for V , then span S = V , and so there exist con-
i
1
n
stants c such that w = c v 1 + · · · + c v n .
i
Suppose there exists a second set of constants d such that
n
1
w = d v 1 + · · · + d v n .
Then

= w − w
0 V
1 n 1 n
= c v 1 + · · · + c v n − d v 1 − · · · − d v n
1
n
1
n
= (c − d )v 1 + · · · + (c − d )v n .
i
i
If it occurs exactly once that c 6= d , then the equation reduces to 0 =
i
i
(c − d )v i , which is a contradiction since the vectors v i are assumed to be
non-zero.
i
i
If we have more than one i for which c 6= d , we can use this last equation
to write one of the vectors in S as a linear combination of other vectors in S,
which contradicts the assumption that S is linearly independent. Then for
i
i
every i, c = d .
Proof Explanation

Remark This theorem is the one that makes bases so useful–they allow us to convert
abstract vectors into column vectors. By ordering the set S we obtain B = (v 1 , . . . , v n )
and can write
 1   1 
c c
. . .
 .   . 
w = (v 1 , . . . , v n )  .  =  . 
c n c n
B
Remember that in general it makes no sense to drop the subscript B on the column
vector on the right–most vector spaces are not made from columns of numbers!

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