Our goal in this section is to show that every linearly independent sequence in a finite-dimensional vector space can be extended, by adding some more vectors to the sequence, to a basis.
Suppose is a linearly independent sequence in a vector space , and . Then implies is linearly independent.
We prove the contrapositive, which is that if is linearly dependent then .
Suppose is linearly dependent. There are scalars , not all of which are zero, such that
can’t be zero, for then this equation would say that was linearly dependent. Therefore we can rearrange to get
as required. ∎
Let be finite-dimensional and let be linearly independent. Then there is a basis of containing .
: Let . Since is finite-dimensional there are elements of that span .
Define a sequence of sequences of elements of as follows: , and for ,
Here just means take the sequence and add on to the end.
Note that in either case , and also that .
Each sequence is linearly independent by the extension lemma, Lemma 4.12.1 and in particular is linearly independent. Furthermore contains the spanning sequence because for each we have , so since subspaces are closed under taking linear combinations, . Therefore is a basis containing . This completes the proof. ∎
As a corollary, we can prove that every finite-dimensional vector space has a basis. Start with any nonzero vector you like — this forms a linearly independent sequence of length 1. The above result lets us extend that to a basis, and in particular, a basis exists.
Consider the sequence of elements where , of the vector space of all width 4 row vectors with real number entries. It’s easy to check that they are linearly independent. We are going to use the procedure above, together with the spanning sequence
of to produce a basis of containing .
We begin with the sequence . To find we have to determine if . It isn’t (to see this, show that the system of linear equations
has no solutions), so is with added, which is .
To find we have to determine if . It is, because
so is the same as .
To find we have to determine if . It is, because
so is the same as .
Finally to find we have to determine if . It is not (no linear combination of can have a nonzero entry in the last position), so is with added. We have run out of s, so is the required basis containing .