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Math Help - Question:Every other subspace of V that contains v1,v2,...,vr must contain W?How?

  1. #1
    Senior Member x3bnm's Avatar
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    Question:Every other subspace of V that contains v1,v2,...,vr must contain W?How?

    I'm having trouble understanding the following proof.

    If v_1 , v_2 , ..., v_r are vectors in a vector space V, then:
    (a) The set W of all linear combinations of v_1 , v_2 , ..., v_r is a subspace of V.
    (b) W is the smallest subspace of V that contains v_1 , v_2 , ..., v_r in the sense that every other subspace of V that contains v_1 , v_2 , ..., v_r must contain W.

    I understand how to prove (a).
    Now I have questions regarding proof (b):

    The proof of (b) is given like this:
    Each vector v_i is a linear combination of v_1 , v_2 , ..., v_r since we can write
    v_i = 0v_1 + 0v_2+... + 1v_i+...+0v_r

    Therefore, the subspace W contains each of the vectors v_1 , v_2 , ..., v_r.

    Let W' be any other subspace that contains v_1 , v_2 , ..., v_r

    Since W' is closed under addition and scalar multiplication, it must contain all linear combinations of v_1 , v_2 , ..., v_r. Thus W' contains each vector of W


    ( My Question ) How does the author deduce that W' contains all linear combinations of v_1 , v_2 , ..., v_r in the last sentence of above proof?

    How does the statement that " W' is closed under addition and scalar multiplication" make the statement that " W' contains all linear combinations of v_1 , v_2 , ..., v_r'' true?
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  2. #2
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    Quote Originally Posted by x3bnm View Post
    How does the statement that " W' is closed under addition and scalar multiplication" make the statement that " W' contains all linear combinations of v_1 , v_2 , ..., v_r'' true?
    Well we know that W' is a subspace.
    It contains each v_k so it contains any linear combination of those.
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  3. #3
    Senior Member x3bnm's Avatar
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    Quote Originally Posted by Plato View Post
    Well we know that W' is a subspace.
    It contains each v_k so it contains any linear combination of those.
    First of all thanks for replying.

    But still I'm not understanding. Sorry about that.

    How does being a subspace that contains v_k make the subspace to contain linear combination of those? I don't understand. Can you explain?
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    a non-empty subset W of V is a subspace if and only if for all w1,w2 in W, and a in F, w1+w2 and aw1 are in W.

    these properties are called "closure of vector addition and scalar multiplication" and are enough to ensure that W is a

    full-fledged vector space in its own right (using the same vector addition and scalar multiplication as in V).

    so if {v1,v2,...,vr} are all in W, then so are the scalar multiples a1v1, a2v2,....arvr and thus also the sum:

    a1v1 + a2v2 +....+ arvr, because of the closure properties listed above (properties any subspace of V has.

    think about this: if W is to be a subspace, then it has to satisfy all of the vector space axioms, or else it

    wouldn't be a subspace, just a subset).
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    Senior Member x3bnm's Avatar
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    Quote Originally Posted by Deveno View Post
    a non-empty subset W of V is a subspace if and only if for all w1,w2 in W, and a in F, w1+w2 and aw1 are in W.

    these properties are called "closure of vector addition and scalar multiplication" and are enough to ensure that W is a

    full-fledged vector space in its own right (using the same vector addition and scalar multiplication as in V).

    so if {v1,v2,...,vr} are all in W, then so are the scalar multiples a1v1, a2v2,....arvr and thus also the sum:

    a1v1 + a2v2 +....+ arvr, because of the closure properties listed above (properties any subspace of V has.

    think about this: if W is to be a subspace, then it has to satisfy all of the vector space axioms, or else it

    wouldn't be a subspace, just a subset).
    Thank you Deveno for the explanation. I understand now.
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    Senior Member x3bnm's Avatar
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    Quote Originally Posted by Deveno View Post
    a non-empty subset W of V is a subspace if and only if for all w1,w2 in W, and a in F, w1+w2 and aw1 are in W.

    these properties are called "closure of vector addition and scalar multiplication" and are enough to ensure that W is a

    full-fledged vector space in its own right (using the same vector addition and scalar multiplication as in V).

    so if {v1,v2,...,vr} are all in W, then so are the scalar multiples a1v1, a2v2,....arvr and thus also the sum:

    a1v1 + a2v2 +....+ arvr, because of the closure properties listed above (properties any subspace of V has.

    think about this: if W is to be a subspace, then it has to satisfy all of the vector space axioms, or else it

    wouldn't be a subspace, just a subset).
    Sorry for reopening this thread. Something is not clear to me.

    According to definition of subspace W is a subspace of V if and only if:

    a) If u and v are vectors in W then u + v is in W.
    b) If k is any scalar and u is any vector in W, then ku is in W.

    So in the above proof v_1, v_2,...,v_3 is in W':

    u = (a_1 * v_1) + (a_2 * v_2) + .... + (a_r * v_r)
    v = (b_1 * v_1) + (b_2 * v_2) + .... + (b_r * v_r)

    so u + v = (a_1 + b_1)v_1 + (a_2 + b_2)v_2 +...+(a_r + b_r)v_r
    which is in W'.

    Deveno did you mean this that u + v is in W' when you said "so if {v1,v2,...,vr} are all in W, then so are the scalar multiples a1v1, a2v2,....arvr and thus also the sum: a1v1 + a2v2 +....+ arvr, because of the closure properties listed above" ?
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  7. #7
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    you just use the properties of closure repeatedly.

    v1 and v2 are in W, so so are a1v1 and a2v2 (a1 and a2 are both scalars, and v1,v2 are both vectors in W). this is property (b) used twice.

    since we have established that both a1v1 and a2v2 are in W, then a1v1 + a2v2 is in W, by property (a).

    again, by property (b), since v3 is in W, a3v3 is in W. so then, by property (a) again:

    a1v1 + a2v2 + a3v3 = (a1v1 + a2v2) + a3v3 is in W.

    we can keep doing this, adding each term ajvj to:

    a1v1 + a2v2 +...+ a(j-1)v(j-1) until we reach arvr.
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  8. #8
    Senior Member x3bnm's Avatar
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    Quote Originally Posted by Deveno View Post
    you just use the properties of closure repeatedly.

    v1 and v2 are in W, so so are a1v1 and a2v2 (a1 and a2 are both scalars, and v1,v2 are both vectors in W). this is property (b) used twice.

    since we have established that both a1v1 and a2v2 are in W, then a1v1 + a2v2 is in W, by property (a).

    again, by property (b), since v3 is in W, a3v3 is in W. so then, by property (a) again:

    a1v1 + a2v2 + a3v3 = (a1v1 + a2v2) + a3v3 is in W.

    we can keep doing this, adding each term ajvj to:

    a1v1 + a2v2 +...+ a(j-1)v(j-1) until we reach arvr.
    Thank you Deveno.
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