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Linear Algebra Proof

  1. Jul 18, 2008 #1

    danago

    User Avatar
    Gold Member

    Prove that if U and V are subspaces of Rn, then so is [tex]U \cap V[/tex]. Show by example that [tex]U \cup V[/tex] is not necessarily a subspace of Rn.

    For the first part i can see that elements of [tex]U \cap V[/tex] are elements of both subspaces U and V. If i take any two element from [tex]U \cap V[/tex] and form some linear combination, then it must lie within U (since the elements are in U and U is a subspace) and it must also lie within V (since the elements are in V and V is a subspace), hence, [tex]U \cap V[/tex] is also a subspace.

    Thats what i would have called informal reasoning. How can i actually prove that this is the case? Is my reasoning enough to stand as a solid proof?

    For the second part i didnt have any troubles. I just defined:

    [tex]
    \begin{array}{l}
    U = \{ k\left( {\begin{array}{*{20}c}
    1 \\
    1 \\
    \end{array}} \right)|k \in R\} \\
    V = \{ k\left( {\begin{array}{*{20}c}
    1 \\
    0 \\
    \end{array}} \right)|k \in R\} \\
    U \cup V = \{ a\left( {\begin{array}{*{20}c}
    1 \\
    1 \\
    \end{array}} \right),b\left( {\begin{array}{*{20}c}
    1 \\
    0 \\
    \end{array}} \right)|a,b \in R\} \\
    \end{array}
    [/tex]

    I then chose two elements from the union of U and V and formed a linear combination which was not an element of the union of U and V, hence U and V is not a subspace.

    [tex]
    \begin{array}{l}
    \left( {\begin{array}{*{20}c}
    0 \\
    1 \\
    \end{array}} \right),\left( {\begin{array}{*{20}c}
    1 \\
    1 \\
    \end{array}} \right) \in U \cup V \\
    \left( {\begin{array}{*{20}c}
    0 \\
    1 \\
    \end{array}} \right) + \left( {\begin{array}{*{20}c}
    1 \\
    1 \\
    \end{array}} \right) = \left( {\begin{array}{*{20}c}
    1 \\
    2 \\
    \end{array}} \right) \notin U \cup V \\
    \end{array}
    [/tex]

    I think thats the way to go about it, so its really just the first part i needed a bit of help with.

    Thanks,
    Dan.
     
  2. jcsd
  3. Jul 18, 2008 #2

    Defennder

    User Avatar
    Homework Helper

    Well you can write it in this form:

    Suppose [tex]\vec{u_1},\vec{u_2} \in U \cap V[/tex].

    Clearly that implies that [tex]\vec{u_1} \ \mbox{and} \ \vec{u_2} \in U[/tex] and this in turn implies that for any [tex]a,b \in \Re , a\vec{u_1} + b\vec{u_2} \in U[/tex] since U is a subspace of R^n. What can you say about u1,u2 in relation to V now? And how would you relate this to [tex]U \cap V[/tex]?
     
  4. Jul 18, 2008 #3
    Yes. It is a perfectly good formal proof. In fact I would much rather see a simple explanation like that than some string of symbols like u,v,a,b with decoration.
     
  5. Jul 18, 2008 #4

    danago

    User Avatar
    Gold Member

    cheers for the replies guys. All pretty clear now :)
     
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