Basis
Definition of basis:
The vectors in a vector space V are said to form a basis of V if
(a) span V (i.e., ).
(b) are linearly independent.
Example:
. Are and a basis in ?
[solution:]
and form a basis in since
(a) (see the example in the previous section).
(b) and are linearly independent (also see the example in the previous section).
Example:
. Are and a basis in ?
[solution:]
and are not a basis of since and are linearly dependent,
.
Note that .
Example:
. Are and a basis in ?
[solution:]
andare not a basis in since and are linearly independent,
.
Example:
Let
.
Are S a basis in ?
[solution:]
(a)
For any vector , there exist real numbers such that
.
we need to solve for the linear system
.
The solution is
.
Thus,
.
That is, every vector in can be a linear combination of and .
(b)
Since
,
are linearly independent.
By (a) and (b), are a basis of .
Important result:
If is a basis for a vector space V, then every vector in V can be written in an unique way as a linear combination of the vectors in S.
Example:
. S is a basis of . Then, for any vector ,
is uniquely determined.
Important result:
Let be a set of nonzero vectors in a vector space V and let . Then, some subset of S is a basis of W.
How to find a basis (subset of S) of W:
There are two methods:
Method 1:
The procedure based on the proof of the above important result.
Method 2:
Step 1: Form equation
.
Step 2: Construct the augmented matrix associated with the equation in step 1 and transform this augmented matrix to the reduced row echelon form.
Step 3: The vectors corresponding to the columns containing the leading 1’s form a basis. For example, if and the reduced row echelon matrix is
,
then the 1’st, the 3’nd, and the 4’th columns contain a leading 1 and thus
are a basis of .
Example:
Let
and . Please find subsets of S which form a basis of .
[solution:]
Method1:
We first check if and are linearly independent. Since they are linearly independent, we continue to check if , and are linearly independent. Since
,
we delete from S and form a new set , . Then, we continue to check if , and are linearly independent.They are linearly independent. Thus, we finally check if , and are linearly independent.Since
,
we delete from and form a new set , . Therefore,
is the subset of S which form a basis of form a basis of.
Method 2:
Step 1:
The equation is
.
Step 2:
The augmented matrix and its reduced row echelon matrix is
.
The 1’st, the 2’nd and 4’th columns contain the leading 1’s. Thus,
forms a basis.
Important result:
Let be a basis for a vector space V and let is a linear independentset of vectors in V. Then, .
Corollary:
Let and be two bases for a vector space V. Then, .
Note:
For a vector space V, there are infinite bases. But the number of vectors in two different bases are the same.
Example:
For the vector space ,
is a basis for (see the previous example). Also,
is basis for .
There are 3 vectors in both S and T.
1