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Math Help - Help on some Linear Algebra - subspaces

  1. #1
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    Help on some Linear Algebra - subspaces

    Mainly I just want verification if this is correct or otherwise.

    Consider two subspaces V and W of R_n.
    Is V U W necessarily a subspace of R? Is the intersect of V and W a subspace?

    Yes to both?
    If a vector x is an element of V intersect W, then it is an element of V, then therefore all linear combinations are also in V. It is also in W, and therefore all linear combinations are in W as well. Thus all linear combinations are in V intersect W. The other one would use similar logic.

    ALSO

    Consider vectors v1, v2... v_m in |Rn. Is the span of the vectors necessarily a subspace of Rn?

    I think this is true but how would I justify it?
    Say that span is all linear combinations, and then c can be 0 so that fulfills one requirement.
    Then say that all scalar multiples of any vectors with real numbers will result in vectors of real numbers?
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    Quote Originally Posted by Lord Voldemort View Post
    Mainly I just want verification if this is correct or otherwise.

    Consider two subspaces V and W of R_n.
    Is V U W necessarily a subspace of R? Is the intersect of V and W a subspace?

    Yes to both?
    If a vector x is an element of V intersect W, then it is an element of V, then therefore all linear combinations are also in V. It is also in W, and therefore all linear combinations are in W as well. Thus all linear combinations are in V intersect W. The other one would use similar logic.

    ALSO

    Consider vectors v1, v2... v_m in |Rn. Is the span of the vectors necessarily a subspace of Rn?

    I think this is true but how would I justify it?
    Say that span is all linear combinations, and then c can be 0 so that fulfills one requirement.
    Then say that all scalar multiples of any vectors with real numbers will result in vectors of real numbers?
    If the vectors in V and W are subspaces of R^n, the union of of course will be subspace.

    The intersection of V and W could be empty. Is an empty set a subspace? I am not too sure of that.
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  3. #3
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    Thanks!
    V intersect W is not necessarily a subspace, since the empty set does not contain 0.
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    Quote Originally Posted by Lord Voldemort View Post
    Thanks!
    V intersect W is not necessarily a subspace, since the empty set does not contain 0.
    I was to lazy to look up anything so goodjob.
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  5. #5
    A Plied Mathematician
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    If V and W are both subspaces, they both contain the zero vector, and hence the intersection is nonempty.
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  6. #6
    Behold, the power of SARDINES!
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    Quote Originally Posted by dwsmith View Post
    If the vectors in V and W are subspaces of R^n, the union of of course will be subspace.

    The intersection of V and W could be empty. Is an empty set a subspace? I am not too sure of that.
    I have to disagee. The union of two subspaces is, NOT in general, a subspace. Here is a very simple counter example.

    Lets consider \mathbb{R}^2

    Now consider the subspace spanned by V=\vec{i} (the x axis)

    and the subspace spanned by W=\vec{j} (the y axis)

    Now the Union of these two subspaces is both coordinate axes. This is not a vector space at \displaystyle \vec{i}+\vec{j} \notin V \cup U

    What is true is if you take the sum V+W=\{v+w,v\in V w \in W\} this is a subspace.
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    Quote Originally Posted by TheEmptySet View Post
    I have to disagee. The union of two subspaces is, NOT in general, a subspace. Here is a very simple counter example.

    Lets consider \mathbb{R}^2

    Now consider the subspace spanned by V=\vec{i} (the x axis)

    and the subspace spanned by W=\vec{j} (the y axis)

    Now the Union of these two subspaces is both coordinate axes. This is not a vector space at \displaystyle \vec{i}+\vec{j} \notin V \cup U

    What is true is if you take the sum V+W=\{v+w,v\in V w \in W\} this is a subspace.
    In fact, if U,W\leqslant V then U\cup W\leqslant V if and only if U\subseteq W or W\subseteq U. Indeed, suppose neither containment held, then we may choose u\in U-W and w\in W-U. A quick check then shows that u+w\notin U since otherwise (u+w)-u=w\in U contrary to construction. Similarly, u+w\notin W. Thus, u+w\notin U\cup W etc. In fact, this is really just a theorem about the underlying group structure of the vector spaces.
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by TheEmptySet View Post

    What is true is if you take the sum V+W=\{v+w,v\in V w \in W\} this is a subspace.
    It's also interesting to note that this is not strange since, if \mathcal{V} is the ambient space, \displaystyle V+W=\bigcap_{\substack{U\leqslant \mathcal{V}\\ V\cup W\subseteq U}}U
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