About
The set of all linear combinations of some vectors v1,…,vn is called the span of these vectors and contains always the origin.
Example: Let V = Span {[0, 0, 1], [2, 0, 1], [4, 1, 2]}. A vector belongs to V when you can write it as a linear combination of the generators of V.
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Definition
The set of all linear combinations of some vectors v1,…,vn is called the span of these vectors and contains always the origin.
Span delim{lbrace}v_1_..._v_n{rbrace}.How to
know if a vector is in the span
How to know if a vector is in the Span {v1, . . . , vn} ?
Let <math> A = \begin{bmatrix} \begin{array}{r|r|r} && \\ v_1 & \dots & v_n \\ && \end{array} \end{bmatrix} </math>
Using the linear-combinations interpretation of matrix-vector multiplication, a vector x in Span {v1, . . . , vn} can be written Ax.
Thus testing if b is in Span {v1, . . . , vn} is equivalent to testing if the matrix equation Ax = b has a solution.
Example
Span {}
The Span {} have only one vector: the zero vector
Span {[1, 1], [0, 1]} over gf2
The Span {[1, 1], [0, 1]} over the field GF(2) is composed of the following linear combinations:
0 [1, 1] + 0 [0, 1] = [0, 0]
0 [1, 1] + 1 [0, 1] = [0, 1]
1 [1, 1] + 0 [0, 1] = [1, 1]
1 [1, 1] + 1 [0, 1] = [1, 0]
Thus there are four vectors in the span.
Span {[2, 3]} over <math>\mathbb{R}</math>
The Span {[2, 3]} over <math>\mathbb{R}</math> contains an infinite number of vectors. They forms the line through the origin and (2, 3).
delim{lbrace}{alpha {delim{[}{2,3}{]}} : alpha in bbR }{rbrace}Span of two vectors
The span of two vectors is a plane containing the origin.
Span in another Span
[3, 0, 0], [0, 2, 0], and [0, 0, 1] are in Span {[1, 0, 0], [1, 1, 0], [1, 1, 1]}:
[3, 0, 0] = 3 [1, 0, 0]
[0, 2, 0] = 2 [1, 0, 0] + 2 [1, 1, 0]
[0, 0, 1] = - 1 [1, 0, 0] - 1 [1, 1, 0] + 1 [1, 1, 1]
Dimension
The span of k vectors is not always k-dimensional
- Span {[0, 0]} is 0-dimensional.
- Span {[1, 3], [2, 6]} is 1-dimensional as [1, 3] = 1/2 x [2, 6]
- Span {[1, 0, 0], [0, 1, 0], [1, 1, 0]} is 2-dimensional as [1, 0, 0] + [0, 1, 0] = [1, 1, 0]
To predict the dimensionality of the span of some vectors, compute the rank of the set of vectors.
Exchange Lemma
Suppose S is a set of vectors and A is a subset of S. Suppose z is a vector in Span S such that <math>A \cup \{z\}</math> is linearly independent. Then there is a vector <math>w \in S − A</math> such that: <MATH> Span S = Span (S \cup \{z\} − \{w\}) </MATH>