Vector Spaces
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Theorem (Vector spaces)
S is a subspace of V. To prove this we have to show that if[Equation goes here - download the original to see it.]then[Equation goes here - download the original to see it.]and if [Equation goes here - download the original to see it.]and[Equation goes here - download the original to see it.]then[Equation goes here - download the original to see it.]. (1) Let [Equation goes here - download the original to see it.]S then [Equation goes here - download the original to see it.] for some[Equation goes here - download the original to see it.]since [Equation goes here - download the original to see it.]are linear combinations of [Equation goes here - download the original to see it.] Then [Equation goes here - download the original to see it.] which is a higher combination of the [Equation goes here - download the original to see it.]; hence[Equation goes here - download the original to see it.] (ii) [Equation goes here - download the original to see it.] If [Equation goes here - download the original to see it.]is a set of vectors the span of this set is denoted by [Equation goes here - download the original to see it.] Example Is the vector[Equation goes here - download the original to see it.]in the subspace spanned by the vectors [Equation goes here - download the original to see it.] Solution If[Equation goes here - download the original to see it.]is in the span[Equation goes here - download the original to see it.]then for some [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] Hence [Equation goes here - download the original to see it.] We need to check this solution for consistency. (2.) gives ; (3.) gives[Equation goes here - download the original to see it.] substituting in (1.) gives [Equation goes here - download the original to see it.] So the solution is consistent. If[Equation goes here - download the original to see it.]was not in the span[Equation goes here - download the original to see it.]then we would not be able to find a consistent choice for[Equation goes here - download the original to see it.]and[Equation goes here - download the original to see it.]. When one vector is in the span of a set of other vectors then it is said to be a linear combination of some of those vectors. That is suppose[Equation goes here - download the original to see it.]was in the span[Equation goes here - download the original to see it.], then, for some[Equation goes here - download the original to see it. [Equation goes here - download the original to see it.]If[Equation goes here - download the original to see it.]does not lie in[Equation goes here - download the original to see it.]then[Equation goes here - download the original to see it.]is linearly independent of[Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.]. A finite set of vectors[Equation goes here - download the original to see it.] is linearly independent if [Equation goes here - download the original to see it.] implies[Equation goes here - download the original to see it.] Otherwise, the set is linearly dependent. A basis for a set V is a linearly independent set of vectors that span V. This means that every vector in V can be written in only one way as a linear combination of the vectors in the spanning set. Example: Prove that[Equation goes here - download the original to see it.]is a basis for[Equation goes here - download the original to see it.]. Solution Any vector in[Equation goes here - download the original to see it.]takes the form[Equation goes here - download the original to see it.] where[Equation goes here - download the original to see it.]. Then [Equation goes here - download the original to see it.] so V spans[Equation goes here - download the original to see it.]. We need to show that V is a set of linearly independent vectors. Suppose [Equation goes here - download the original to see it.] then [Equation goes here - download the original to see it.] so V is a linearly independent set of vectors; hence V is a basis for[Equation goes here - download the original to see it.]. The set[Equation goes here - download the original to see it.]is not the only basis for[Equation goes here - download the original to see it.]. Any set of three linearly independent vectors in[Equation goes here - download the original to see it.]. For example, another basis is provided by [Equation goes here - download the original to see it.]
S is a subspace of V. To prove this we have to show that if[Equation goes here - download the original to see it.]then[Equation goes here - download the original to see it.]and if [Equation goes here - download the original to see it.]and[Equation goes here - download the original to see it.]then[Equation goes here - download the original to see it.]. (1) Let [Equation goes here - download the original to see it.]S then [Equation goes here - download the original to see it.] for some[Equation goes here - download the original to see it.]since [Equation goes here - download the original to see it.]are linear combinations of [Equation goes here - download the original to see it.] Then [Equation goes here - download the original to see it.] which is a higher combination of the [Equation goes here - download the original to see it.]; hence[Equation goes here - download the original to see it.] (ii) [Equation goes here - download the original to see it.] If [Equation goes here - download the original to see it.]is a set of vectors the span of this set is denoted by [Equation goes here - download the original to see it.] Example Is the vector[Equation goes here - download the original to see it.]in the subspace spanned by the vectors [Equation goes here - download the original to see it.] Solution If[Equation goes here - download the original to see it.]is in the span[Equation goes here - download the original to see it.]then for some [Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.] Hence [Equation goes here - download the original to see it.] We need to check this solution for consistency. (2.) gives ; (3.) gives[Equation goes here - download the original to see it.] substituting in (1.) gives [Equation goes here - download the original to see it.] So the solution is consistent. If[Equation goes here - download the original to see it.]was not in the span[Equation goes here - download the original to see it.]then we would not be able to find a consistent choice for[Equation goes here - download the original to see it.]and[Equation goes here - download the original to see it.]. When one vector is in the span of a set of other vectors then it is said to be a linear combination of some of those vectors. That is suppose[Equation goes here - download the original to see it.]was in the span[Equation goes here - download the original to see it.], then, for some[Equation goes here - download the original to see it. [Equation goes here - download the original to see it.]If[Equation goes here - download the original to see it.]does not lie in[Equation goes here - download the original to see it.]then[Equation goes here - download the original to see it.]is linearly independent of[Equation goes here - download the original to see it.] [Equation goes here - download the original to see it.]. A finite set of vectors[Equation goes here - download the original to see it.] is linearly independent if [Equation goes here - download the original to see it.] implies[Equation goes here - download the original to see it.] Otherwise, the set is linearly dependent. A basis for a set V is a linearly independent set of vectors that span V. This means that every vector in V can be written in only one way as a linear combination of the vectors in the spanning set. Example: Prove that[Equation goes here - download the original to see it.]is a basis for[Equation goes here - download the original to see it.]. Solution Any vector in[Equation goes here - download the original to see it.]takes the form[Equation goes here - download the original to see it.] where[Equation goes here - download the original to see it.]. Then [Equation goes here - download the original to see it.] so V spans[Equation goes here - download the original to see it.]. We need to show that V is a set of linearly independent vectors. Suppose [Equation goes here - download the original to see it.] then [Equation goes here - download the original to see it.] so V is a linearly independent set of vectors; hence V is a basis for[Equation goes here - download the original to see it.]. The set[Equation goes here - download the original to see it.]is not the only basis for[Equation goes here - download the original to see it.]. Any set of three linearly independent vectors in[Equation goes here - download the original to see it.]. For example, another basis is provided by [Equation goes here - download the original to see it.]
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Contents of
Vector Spaces
1 Vector Spaces
2 Linear polynomials
3 Complex numbers
4 Quadratic polynomials
5 Continuous Functions
6 Infinite sequences
7 Vector space axioms
8 Properties of vector spaces
9 Subspaces of vector spaces
10 Basis and dimension
11 Span
12 Theorem (Vector spaces)
13 Vectors, dimension and bases
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