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Math Help - Confusing subspace problem

  1. #1
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    Confusing subspace problem

    The question:

    Show that T = { \sum_{i=1}^4 \lambda _i v_i  : \lambda _i \in \mathbb{R}, 1 \le i \le 4}, where v_1, v_2, v_3, v_4 are given fixed vectors in \mathbb{R}^3 is either a subspace or not a subspace of \mathbb{R}^3.

    I'm not sure what they mean by "given fixed vectors" means, or how to use them in my reasoning. I'm also a little confused as to how I can check the Subspace Theorem when there's summation involved. Any assistance would be great.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Glitch View Post
    The question:

    Show that T = { \sum_{i=1}^4 \lambda _i v_i  : \lambda _i \in \mathbb{R}, 1 \le i \le 4}, where v_1, v_2, v_3, v_4 are given fixed vectors in \mathbb{R}^3 is either a subspace or not a subspace of \mathbb{R}^3.

    I'm not sure what they mean by "given fixed vectors" means, or how to use them in my reasoning. I'm also a little confused as to how I can check the Subspace Theorem when there's summation involved. Any assistance would be great.
    So I want you to try this on your own (and report back if you haven any trouble). An example of what the question is asking you to figure out is given the four 'fixed' (not changing) vectors \left\{\begin{bmatrix}1\\ 0\\ 3\\ 4\end{bmatrix},\begin{bmatrix}0\\ 3\\ 28\\ 7\end{bmatrix},\begin{bmatrix}\pi\\ 37\\ e\\ 2\end{bmatrix},\begin{bmatrix}0\\ 0\\ 0\\ 1\end{bmatrix}\right\} you can, for each quadruple of 'fixed' (given, not changin) real numbers r_1,r_2,r_3,r_4 you can make a new vector v_{r_1,r_2,r_3,r_4}=r_1\begin{bmatrix}1\\ 0\\ 3\\ 4\end{bmatrix}+r_2\begin{bmatrix}0\\ 3\\ 28\\ 7\end{bmatrix}+r_3\begin{bmatrix}\pi\\ 37\\ e\\ 2\end{bmatrix}+r_4\begin{bmatrix}0\\ 0\\ 0\\ 1\end{bmatrix}=\begin{bmatrix}r_1+r_3\pi\\ 3r_2+37r_3\\ 3r_1+28r_2+er_3\\ 4r_1+7r_2+2r_3+r_4\end{bmatrix}. You can then consider the set \left\{v_{r_1,r_2,r_3,r_4}:r_1,r_2,r_3,r_4\in\math  bb{R}\right\}. Is that set closed under linear combinations?
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  3. #3
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    I think I see what you're saying. This is what I have:

    i) Let v_i \in \mathbb{R}^3
    Let \lambda_i = 0
    \lambda_i v_i = 0v_i = 0
    Therefore contains zero vector

    ii) \sum_{i=1}^4\lambda_i v_i = \lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 + \lambda_4 v_4
    Let \alpha \in \mathbb{R}

    \alpha \sum_{i=1}^4\lambda_i v_i = \alpha(\lambda_1 v_1 + \lambda_2 v_2 + \lambda_3 v_3 + \lambda_4 v_4)

    \alpha \sum_{i=1}^4\lambda_i v_i = \alpha\lambda_1 v_1 + \alpha\lambda_2 v_2 + \alpha\lambda_3 v_3 + \alpha\lambda_4 v_4

    = \sum_{i=1}^4\alpha\lambda_i v_i
    = \sum_{i=1}^4(\alpha\lambda_i)v_i
    Since \alpha\lambda_i \in \mathbb{R}, set T is closed under scalar multiplication

    iii) Let a, b \in \mathbb{R}^3
    \sum_{i=1}^4\lambda_i a_i + \sum_{i=1}^4\lambda_i b_i
    = \sum_{i=1}^4\lambda_i a_i + \lambda_i b_i
    = \sum_{i=1}^4\lambda_i(a_i + b_i)
    Since a_i + b_i \in \mathbb{R}^3, set T is closed under vector addition.

    Therefore, T is a subspace of \mathbb{R}^3

    is this proof correct? Thanks.
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