Page 216 - 35Linear Algebra

216 Basis and Dimension

vectors in the collection S 1 ; rewriting v i as such allows us to express u as
a linear combination of the vectors in S 1 . Thus S 1 is a basis of V with n
vectors.
We can now iterate this process, replacing one of the v i in S 1 with w 2 ,
and so on. If m ≤ n, this process ends with the set S m = {w 1 , . . . , w m ,
}, which is fine.
v i 1 , . . . , v i n−m
Otherwise, we have m > n, and the set S n = {w 1 , . . . , w n } is a basis

for V . But we still have some vector w n+1 in T that is not in S n . Since S n
is a basis, we can write w n+1 as a combination of the vectors in S n , which
contradicts the linear independence of the set T. Then it must be the case
that m ≤ n, as desired.

Worked Example

Corollary 11.0.3. For a finite-dimensional vector space V , any two bases
for V have the same number of vectors.

Proof. Let S and T be two bases for V . Then both are linearly independent
sets that span V . Suppose S has n vectors and T has m vectors. Then by
the previous lemma, we have that m ≤ n. But (exchanging the roles of S
and T in application of the lemma) we also see that n ≤ m. Then m = n,
as desired.

Reading homework: problem 2

n
11.1 Bases in R .

In review question 2, chapter 10 you checked that

     
 1 0

 0 
0
1
0
 
     
n
R = span       ,
. . .
 . ,  . , . . . ,  .
 .  .  .
 
 
0 0 1
 
and that this set of vectors is linearly independent. (If you didn’t do that
problem, check this before reading any further!) So this set of vectors is
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