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Thread: Show a set is not a subspace of R^4

  1. #1
    Member
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    Jan 2011
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    87

    Show a set is not a subspace of R^4

    Hi all,

    Having a problem trying to show that the set $\displaystyle
    \begin{bmatrix} 3a-b\\ ab \\ a+b\\ b+2a \end{bmatrix}
    $
    where a and b are real numbers, is not a subspace in R^4..
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  2. #2
    Senior Member
    Joined
    Feb 2008
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    Re: Show a set is not a subspace of R^4

    Let $\displaystyle a=1,b=2$ and consider the vector

    $\displaystyle 2\left[\begin{array}{c}3a-b\\ab\\a+b\\2a+b\end{array}\right]=2\left[\begin{array}{c}1\\2\\3\\4\end{array}\right]=\left[\begin{array}{c}2\\4\\6\\8\end{array}\right]$.

    Then $\displaystyle 3a'-b'=2$, $\displaystyle a'b'=4$ and $\displaystyle a'+b'=6$. But this system has no solution:

    $\displaystyle 3a'-b'=2$ implies $\displaystyle b'=3a'-2$. Then $\displaystyle a'+b'=6$ gives us $\displaystyle a'+(3a'-2)=6$ and therefore $\displaystyle a'=2$ and $\displaystyle b'=4$. But then $\displaystyle a'b'=8$, contradicting the condition $\displaystyle a'b'=4$.
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