Determine if the vector is in the subspace of R2 times 3 giv

Determine if the vector is in the subspace of R^2 times 3 given by span {[1 2 1 0 1 3], [0 3 1 -1 1 0]} v = [2 1 1 1 1 2] v is in the given subspace. v is not in the given subspace.

Solution

Ans-

The real vector space R³ has {(1,0,0), (0,1,0), (0,0,1)} as a spanning set. This particular spanning set is also a basis. If (1,0,0) were replaced by (-1,0,0), it would also form thecanonical basis of R³.

Another spanning set for the same space is given by {(1,2,3), (0,1,2), (1,1/2,3), (1,1,1)}, but this set is not a basis, because it is linearly dependent.

The set {(1,0,0), (0,1,0), (1,1,0)} is not a spanning set of R³; instead its span is the space of all vectors in R³ whose last component is zero. That space (the space of all vectors inR³ whose last component is zero) is also spanned by the set {(1,0,0), (0,1,0)}, as (1,1,0) is a linear combination of (1,0,0) and (0,1,0). It does, however, span R².

The set of functions xⁿ where n is a non-negative integer spans the space of polynomials.

Theorems[edit]

Theorem 1: The subspace spanned by a non-empty subset S of a vector space V is the set of all linear combinations of vectors in S.

This theorem is so well known that at times it is referred to as the definition of span of a set.

Theorem 2: Every spanning set S of a vector space V must contain at least as many elements as any linearly independent set of vectors from V.

Theorem 3: Let V be a finite-dimensional vector space. Any set of vectors that spans V can be reduced to a basis for V by discarding vectors if necessary (i.e. if there are linearly dependent vectors in the set). If the axiom of choice holds, this is true without the assumption that V has finite dimension.

This also indicates that a basis is a minimal spanning set when V is finite-dimensional.

Generalizations[edit]

Generalizing the definition of the span of points in space, a subset X of the ground set of a matroid is called a