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Math Help - Proving this is a subspace

  1. #1
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    Proving this is a subspace

    Can anyone help me with this elementary problem? How do I prove the following is a subspace?

    W = {(a1, a2, a3) | 2*a1 - 7*a2 +a3 = 0}

    I know I have to prove its closed under addition and multiplication... But the condition on W is confusing me...
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  2. #2
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    I think I just solved it, actually!

    I let a3 = 7*a2 - 2*a1, and made that the new condition on W.

    I then let a = (a1, a2, 7*a2 - 2*a1) and similarly for b.

    Then I did a + cb = (a1 + c*b1, a2 + c*b2, 7(a2 + c*b2) - 2(a1 + c*b1))

    Which satisfies my new condition on W I believe...

    Can anyone verify this?
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  3. #3
    Super Member Random Variable's Avatar
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     W = ({a_{1},a_{2},-2a_{1}+7a_{2}} )

    Let  X = (x_{1},x_{2},-2x_{1}+7x_{2}) and  Y = (y_{1},y_{2},-2y_{1}+7y_{2}) be in W

    then X+Y= (x_{1}+y_{1},x_{2}+y_{2},-2x_{1}+7x_{2}-2y_{1}+7y_{2})  = \Big (x_{1}+y_{1},x_{2}+y_{2},-2(x_{1}+y_{1})+7(x_{2}+y_{2}) \Big) which is in W

    I'm sure you can then prove that it's also closed under scalar multiplication.
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  4. #4
    Super Member Random Variable's Avatar
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    If you want to prove both at the same time, you have to show that  \alpha X + \beta Y is in  W
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  5. #5
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    A= \begin{bmatrix} 2 & -7 & 3 \end{bmatrix}, \vec{a}=\begin{bmatrix} a_1\\a_2\\a_3 \end{bmatrix}, W=\{A\vec{a}=\vec{0}\}=ker(A)

    Obviously, T(\vec{x})=A\vec{x} is linear transformation
    1.\ \vec{x}_1,\vec{x}_2 \in W, T(\vec{x}_1+\vec{x}_2)=A(\vec{x}_1+\vec{x}_2)=A(\v  ec{x}_1)+A(\vec{x}_2)=\vec{0}+\vec{0}=\vec{0}, so\ \vec{x}_1+\vec{x}_2 \in W
    2.\ k\in R, T(k\vec{x})=A(k\vec{x})=kA\vec{x}=k\vec{0}=\vec{0} , so\ k\vec{x}\in W
    3.\ A\vec{0}=\vec{0}, so\ \vec{0}\in W

    Finally, W is a subspace.
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  6. #6
    Lord of certain Rings
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    Quote Originally Posted by Random Variable View Post
    If you want to prove both at the same time, you have to show that  \alpha X + \beta Y is in  W
    We can save a greek variable

    It suffices to show  \alpha X + Y is in  W
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