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Math Help - Subspace

  1. #1
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    Subspace

    Show that the following set of elements in R^3 form subspaces.

    (a) The set of all (x,y,z) such that x + y + z = 0

    What is the vector i would use:
    (-z-y, -z-x, -y-x)?
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  2. #2
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    Quote Originally Posted by millerst View Post
    Show that the following set of elements in R^3 form subspaces.

    (a) The set of all (x,y,z) such that x + y + z = 0

    What is the vector i would use:
    (-z-y, -z-x, -y-x)?
    hello
    (-y-z,y,z)=y(-1,1,0)+z(-1,0,1)
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  3. #3
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    So:

    <br /> <br />
(-y-z,y,z)+(-y-z,y,z) = (-2y-2z,2y,2z)<br />
    So it is closed under addition.

    q(-y-z,y,z) = (q(-y-Z),qy,qz)
    So it is closed under multiplication.

    (-y-z,y,z)+(-1)(-y-z,y,z) = 0
    So zero vector is in the set?

    Does this prove it? :S
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  4. #4
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    Quote Originally Posted by millerst View Post
    So:

    <br /> <br />
(-y-z,y,z)+(-y-z,y,z) = (-2y-2z,2y,2z)<br />
    So it is closed under addition.

    q(-y-z,y,z) = (q(-y-Z),qy,qz)
    So it is closed under multiplication.

    (-y-z,y,z)+(-1)(-y-z,y,z) = 0
    So zero vector is in the set?

    Does this prove it? :S
    make it easier,
    E=\left \{ (x,y,z)\in \mathbb{R}^3\mid x+y+z=0 \right \} =\left \{ y(-1,1,0)+z(-1,0,1)<br />
\mid y,z\in \mathbb{R }\right \}=\texttt{Span}((-1,1,0),(-1,0,1))
    and that proves it.
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  5. #5
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    Quote Originally Posted by millerst View Post
    So:

    <br /> <br />
(-y-z,y,z)+(-y-z,y,z) = (-2y-2z,2y,2z)<br />
    So it is closed under addition.
    This doesn't prove it is closed under addition, only that it is closed under addition of a vector to itself. You need to show that it is closed under addition of any two vectors:
    (-y-z, y, z)+ (-b-c, b, c)= (-(y+b)-(z+c), y+b, z+c).

    Or you could do it this way: If (a, b, c) and (x, y, z) are in this set then x + y + z = 0 and a+ b+ c= 0.
    Then (a, b, c)+ (x, y, z)= (a+x, b+ y, c+ z) and (a+x)+ (b+y)+ (c+ z)= (a+ b+ c)+ (x+ y+ z)= 0+ 0= 0 so this sum vector satisfies the same condition and is in the subspace.

    q(-y-z,y,z) = (q(-y-Z),qy,qz)
    So it is closed under multiplication.

    (-y-z,y,z)+(-1)(-y-z,y,z) = 0
    So zero vector is in the set?

    Does this prove it? :S
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