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Math Help - Subspaces

  1. #1
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    Subspaces

    I need to prove that these statements are true or false.

    1. It is possible to find a pair of two-dimensional subspaces S and T of R such that S ∩ T = {0}

    2. If S and T are subspaces of a vector space V, then S U T is a subspace of V.

    3. If S and T are subspace of a vector space V, then S ∩ T is a subspace of V.

    I understand that the two closure identities need to be met in order to be a subspace. However the intersection and union properties are throwing me off. Can anyone help me with this? Thank you.
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  2. #2
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    Quote Originally Posted by pakman View Post
    1. It is possible to find a pair of two-dimensional subspaces S and T of R such that S ∩ T = {0}
    Let \{ \bold{a}_1,\bold{a}_2\} be a basis for S. And \{ \bold{b}_1,\bold{b}_2\} be a basis for T. Note that \bold{b}_1 \not \in \text{spam}\{ \bold{a}_1,\bold{a}_2\}=S because S\cap T = \{ \bold{0} \}. Similarly, \bold{b}_2\not \in \text{spam}\{ \bold{a}_1,\bold{a}_2\}. This means the set \{ \bold{a}_1,\bold{a}_2,\bold{b}_1,\bold{b}_2\} is a linearly independent set. That is a contradiction. Why?
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Let \{ \bold{a}_1,\bold{a}_2\} be a basis for S. And \{ \bold{b}_1,\bold{b}_2\} be a basis for T. Note that \bold{b}_1 \not \in \text{spam}\{ \bold{a}_1,\bold{a}_2\}=S because S\cap T = \{ \bold{0} \}. Similarly, \bold{b}_2\not \in \text{spam}\{ \bold{a}_1,\bold{a}_2\}. This means the set \{ \bold{a}_1,\bold{a}_2,\bold{b}_1,\bold{b}_2\} is a linearly independent set. That is a contradiction. Why?
    Because S and T are a linear combination to get 0?
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  4. #4
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    Quote Originally Posted by pakman View Post
    Because S and T are a linear combination to get 0?
    No. Remember the dimension of \mathbb{R}^3 over \mathbb{R} is 3.
    This means that you cannot have a set with more than 3 elements which are linearly independent.

    2. If S and T are subspaces of a vector space V, then S U T is a subspace of V.
    Let S = \{ (a,0,b)|a,b\in \mathbb{R}\} and T = \{ (a,b,0)|a,b\in \mathbb{R}\}.
    Do the following steps.
    1)Show S,T are sub-spaces.
    2)Show S\cup T is not closed, i.e. there are \bold{x},\bold{y}\in S\cup T so that \bold{x}+\bold{y}\not \in S\cup T
    3)Conclude that S\cup T is not a sub-space.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    No. Remember the dimension of \mathbb{R}^3 over \mathbb{R} is 3.
    This means that you cannot have a set with more than 3 elements which are linearly independent.
    That makes sense now, thanks!

    Let S = \{ (a,0,b)|a,b\in \mathbb{R}\} and T = \{ (a,b,0)|a,b\in \mathbb{R}\}.
    Do the following steps.
    1)Show S,T are sub-spaces.
    2)Show S\cup T is not closed, i.e. there are \bold{x},\bold{y}\in S\cup T so that \bold{x}+\bold{y}\not \in S\cup T
    3)Conclude that S\cup T is not a sub-space.
    s = (1,0,1) and u = (1,1,0) then sUv = (2,2,1) which does not belong to either s or u. Does that work?
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  6. #6
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    Quote Originally Posted by pakman View Post
    Does that work?
    Yes.
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