Page 407 - 35Linear Algebra

G.10 Basis and Dimension 407

G.10 Basis and Dimension

Proof Explanation
Lets walk through the proof of theorem 11.0.1. We want to show that for
S = {v 1 , . . . , v n } 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 .
We should remember that since S is a basis for V , we know two things
• V = span S

• v 1 , . . . , v n are linearly independent, which means that whenever we have
1
n
i
a v 1 + . . . + a v n = 0 this implies that a = 0 for all i = 1, . . . , n.
i
This first fact makes it easy to say that there exist constants c such that
1 n 1 n
w = c v 1 + · · · + c v n . What we don’t yet know is that these c , . . . c are unique.
In order to show that these are unique, we will suppose that they are not,
and show that this causes a contradiction. So suppose there exists a second
i
set of constants d such that
1
n
w = d v 1 + · · · + d v n .
i i
For this to be a contradiction we need to have c 6= d for some i. Then look
what happens when we take the difference of these two versions of w:
= w − w
0 V
1
n
n
1
= (c v 1 + · · · + c v n ) − (d v 1 + · · · + d v n )
n
n
1
1
= (c − d )v 1 + · · · + (c − d )v n .
i
i
Since the v i ’s are linearly independent this implies that c − d = 0 for all i,
i
i
this means that we cannot have c 6= d , which is a contradiction.
Worked Example
In this video we will work through an example of how to extend a set of linearly
independent vectors to a basis. For fun, we will take the vector space
5
V = {(x, y, z, w)|x, y, z, w ∈ Z } .
4
This is like four dimensional space R except that the numbers can only be
{0, 1, 2, 3, 4}. This is like bits, but now the rule is
0 = 5 .

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