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Math Help - [SOLVED] Can (m-1)vectors of R^m generate R^(m-1)?

  1. #1
    Lok
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    [SOLVED] Can (m-1)vectors of R^m generate R^(m-1)?

    Hello,

    I'm trying to figure out if (m-1)vectors of R^m can generate R^(m-1)?

    What I've tried to do so far is this:
    Let's take u_1,u_2,u_3 element of R^4 (with m=4)

    By definition (I think)
    u = sum(lambda_i*u_i,k=1,k=n)

    I suppose u = (x,y,z) element of R^3 (m-1), u_1=(1,0,0,0), u_2=(0,1,0,0), u_3=(0,0,1,0)
    A=[u_1 u_2 u_3]=
    1 0 0
    0 1 0
    0 0 1
    0 0 0

    We know A*lamda_matrix=u
    [ 1 0 0 ][lambda_1]=[lambda_1]=?? [x]
    [ 0 1 0 ][lambda_2] [lambda_2] [y]
    [ 0 0 1 ][lambda_3] [lambda_3] [z]
    [ 0 0 0 ]...............=.. [0]
    (4x3 * 3x1 = 4x1)
    (I used ..... so the [0] wouldn't go in the multiplication, forum escapes whitespace)

    So can I say that (lambda_1,lambda_2,lambda_3,0)=(x,y,z)? I would doubt so...

    If (m-1) vectors from R^m can generate R^(m-1) I would like a proof and if not a counter-example.

    Is it the right way to go with this or should I use something else?

    Thanks (:
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  2. #2
    Member Ruun's Avatar
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    Can you generate \mathbb{R}^2 with 2 vectors of \mathbb{R}^3?

    If you can, when and why? Then generalize to your spaces.
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  3. #3
    Lok
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    Well I have no clue

    If (lambda_1,lambda_2,lambda_3,0) can be equal to (x,y,z), then yes. But I would doubt so... I think you can't get R^(m-1) with vectors from R^m since there is one extra coordinate.

    Sound right?
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  4. #4
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    Quote Originally Posted by Lok View Post
    Hello,

    I'm trying to figure out if (m-1)vectors of R^m can generate R^(m-1)?
    No. Because vectors in \mathbb{R}^m has m components. While \mathbb{R}^{m-1} is a set of ordered coordinates that have m-1 components, so the two sets can never be equal. However, it is possible to generate a subspace with m-1 vectors so that this subspace is isomorphism to \mathbb{R}^{m-1}. Just let v_j = (0,...,1,...,0,0) where 1 is in the j-th location and 1\leq j\leq m-1. Then \text{spam}\{v_1,...,v_{m-1}\} would generate a subspace isomorphic to \mathbb{R}^{m-1}. But your problem is asking if they can be equal. In that case the answer is no.
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  5. #5
    Lok
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    Thanks a lot
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