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Math Help - Subspace proof

  1. #1
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    Subspace proof

    The question:
    Show that the set
    S = \{x \in \mathbb{R}^4 | x = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix} for some \lambda \in \mathbb{R}\}
    is a subspace of \mathbb{R}

    My attempt:
    Using Subspace Theorem:

    S clearly contains the zero vector in \mathbb{R}^4 since the set is all solutions to the parametric form of a line through the origin.

    Let x,y \in S
    x + y = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}
    = 2\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}
    Since \lambda \in \mathbb{R} and 2 \in \mathbb{R}, 2\lambda \in \mathbb{R}, so
    = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}
    (I'm unsure of that reasoning)
    Therefore closed under vector addition.

    Let \alpha \in \mathbb{R}, x \in S
    \alphax = \alpha(\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix})
    = \alpha\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}
    = Since \lambda \in \mathbb{R} and \alpha \in \mathbb{R}, \alpha\lambda \in \mathbb{R}, so
    = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}
    Therefore closed under scalar multiplication.

    Is this proof sound? Thanks.
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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 the set
    S = \{x \in \mathbb{R}^4 | x = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix} for some \lambda \in \mathbb{R}\}
    is a subspace of \mathbb{R}

    My attempt:
    Using Subspace Theorem:

    S clearly contains the zero vector in \mathbb{R}^4 since the set is all solutions to the parametric form of a line through the origin.

    Let x,y \in S
    x + y = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}
    = 2\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}
    Since \lambda \in \mathbb{R} and 2 \in \mathbb{R}, 2\lambda \in \mathbb{R}, so
    = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}
    (I'm unsure of that reasoning)
    Therefore closed under vector addition.

    Let \alpha \in \mathbb{R}, x \in S
    \alphax = \alpha(\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix})
    = \alpha\lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}
    = Since \lambda \in \mathbb{R} and \alpha \in \mathbb{R}, \alpha\lambda \in \mathbb{R}, so
    = \lambda\begin{pmatrix} 2 \\ 0 \\ 1 \\ -1 \end{pmatrix}
    Therefore closed under scalar multiplication.

    Is this proof sound? Thanks.
    I think you have a fundamental misunderstanding of the notation. The \lambda is not fixed. So, for example to show that two elements of your set are closed you need to show that \lambda_1 M+\lambda_2 M=\lambda_3 M where M is the vector in the problem.
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  3. #3
    Senior Member
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    Ahh, I see. So for vector addition, it should be:

    x + y = \lambda_1M + \lambda_2M
    = (\lambda_1 + \lambda_2)M
    Since (\lambda_1 + \lambda_2) \in R, S is closed under vector addition.

    Correct?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Glitch View Post
    Ahh, I see. So for vector addition, it should be:

    x + y = \lambda_1M + \lambda_2M
    = (\lambda_1 + \lambda_2)M
    Since (\lambda_1 + \lambda_2) \in R, S is closed under vector addition.

    Correct?
    Exactly!
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  5. #5
    Senior Member
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    Thanks. I knew I was going something silly!
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