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Connection between subspace, span, and basis?

  1. Oct 18, 2015 #1
    I'm stuck on a relation issue if there is a direct relation at all.

    If I were to verify that a subset is a subspace of a vector space V, would it then be correct to check that subset for linear independence to verify that the subset spans the subspace? I'm not sure if I'm following the material quiet correctly.

    I completely understand that if the set is linearly independent it is a basis for V and that if S spans V it is a basis, but I'm unsure of the process connection to any verification of a subspace.

    Thank you.
     
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  3. Oct 18, 2015 #2

    FactChecker

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    As you stated the problem, that would not be necessary. You have a set defined and I assume it is a subset of the vector space. Now you need to show that the set has the properties of a space. Most properties will be inherited from the original vector space and you can just say which properties are inherited (associative, commutative, etc.). Then you need to show that the subset has all the vectors it needs to be a subspace ( 0 in the subset, rV, V1+V2 in the subset for every r∈R and V1, V2 in the subset.)
     
    Last edited: Oct 18, 2015
  4. Oct 18, 2015 #3

    mathwonk

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    the vectors (0,1) and (1,0) form a basis for the plane, and any set that contains these vectors spans the plane. The plane is a subspace.

    In 3 space, the subset (1,1,2) and (1, 2,2) is a basis for a plane in R^3, while not every subset containing these vectors spans exactly that plane, any subset consisting of those vectors and sums of scalar multiples of them does so. E.g. the subset (1,1,2), (1,2,2) and their sum (2,3,4) spans that same plane.
     
    Last edited: Oct 19, 2015
  5. Oct 19, 2015 #4

    Fredrik

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    "The x axis" ##\{(x,0)|x\in\mathbb R\}## is a subspace of "the x-y plane" ##\mathbb R^2##, but it's not a linearly independent set.

    ##\{(1,0,0),(0,1,0)\}## is a linearly independent subset of ##\mathbb R^3##, but it doesn't span ##\mathbb R^3##, so it can't be a basis for ##\mathbb R^3##.

    ##\{x\in\mathbb R^3:|x|<1\}## spans ##\mathbb R^3##, but isn't linearly independent, so it can't be a basis for ##\mathbb R^3##.
     
    Last edited: Oct 19, 2015
  6. Oct 19, 2015 #5

    HallsofIvy

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    (0, 1, 0), (1, 0, 0), and (0, 2, 0) span a subspace of R3, specifically the subspace of all vectors of the form (x, y, 0) for any numbers x and y. But the three vectors are NOT independent so do not form a basis. Either {(0, 1, 0), (1, 0, 0)} or {(0, 2, 0),(1, 0, 0)} is a basis for that two dimensional subspace.
     
  7. Nov 4, 2015 #6
    It seems perhaps your issue is with the definitions. Then span of a set of n vectors, written as [itex]span(\{ v_1, v_2, \dotsm , v_n \})[/itex] is the set of all possible linear combinations of those n vectors. A basis is a collection of linearly independent vectors whose span is a vector space.

    To verify that a subset of V is also a subspace of V, you would need to show that the subset itself satisfies the properties of a vector space (contains 0, closed under vector addition, associative under vector addition, etc. -- there are about 10 of these properties that must be satisfied).

    Do you? If a set S of vectors spans V, S is not necessarily a basis. As an example,

    [itex]span(\{ [1,0], [0,1], [2,3] \}) = \mathbb{R}^2[/itex]

    but clearly { [1,0], [0,1], [2,3] } is not linearly independent. However, note that

    [itex]span(\{ [1,0], [0,1] \}) = \mathbb{R}^2[/itex]

    also, and { [1,0], [0,1] } is a basis (the difference here is that this set is linearly independent). Also, a set of vectors (in V) that is linearly independent is not necessarily a basis of V. For example, [itex]S=\{ [1,0,0], [0,1,0]\} [/itex] is a set of linearly independent vectors in [itex]\mathbb{R}^3[/itex] but S does not form a basis for [itex]\mathbb{R}^3[/itex] -- in fact, span(S) is a basis for a subspace that is isomorphic to [itex]\mathbb{R}^2[/itex]. We need the size of the basis set to be equal to the dimension of the vector space!
     
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