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Math Help - Subspaces (basic)

  1. #1
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    Subspaces (basic)

    I have to find which of these subsets are subspaces of  R^3

    a)  (a,2a,3a) \mid a \epsilon R
    b)  (x,y,z) \mid 2x - 3y + 4z = 1
    c)  (1,a,b) \mid a,b \epsilon R
    d)  (a-2b,b,a+b) \mid a,b \epsilon R
    e)  (x,y,z) \mid 2x - z = 0

    I know that for a subset to be a subspace of a given set, it can't be an empty set, it must be closed under addition and closed under scalar multiplication.
    I can see that  a) is going to be a subspace of  R^3 because if  a is real, then a real number plus a real number will yield a real number. Also, a real number multiplied by a real number will yield a real number.

    But I don't know how to prove it, nor how to analyse any of the others algebraically.

    Thanks in advance,
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  2. #2
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    For the first one: S=\lbrace (a,2a,3a) \mid a \in R \rbrace \subset R^3.

    Closure under vector addition:

    Let \vec{x}=(a_1,2a_1,3a_1), ~\vec{y}= (a_2,2a_2,3a_2)\in S.
    \vec{x}+\vec{y}=(a_1+a_2, 2a_1+2a_2, 3a_1+3a_2)
    Grouping coefficients,
    = (a_1+a_2,2(a_1+a_2), 3(a_1+a_2)) \in S.
    Thus S is closed under vector addition.

    Scalar multiplication can be showed similarly. (Hint, define an \alpha \in R and show that \forall \vec{x} \in S,~ \alpha \vec{x} \in S).
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  3. #3
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    You should also state that the zero vector of R^3 is in your subspace.
    This is usually one of the first checks on makes if a subset is a subspace.
    Because if the zero vector is not there, then you donīt even have to test for vector addition and scalar multiplication.
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  4. #4
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    Quote Originally Posted by scorpion007 View Post
    For the first one: S=\lbrace (a,2a,3a) \mid a \in R \rbrace \subset R^3.

    Closure under vector addition:

    Let \vec{x}=(a_1,2a_1,3a_1), ~\vec{y}= (a_2,2a_2,3a_2)\in S.
    \vec{x}+\vec{y}=(a_1+a_2, 2a_1+2a_2, 3a_1+3a_2)
    Grouping coefficients,
    = (a_1+a_2,2(a_1+a_2), 3(a_1+a_2)) \in S.
    Thus S is closed under vector addition.

    Scalar multiplication can be showed similarly. (Hint, define an \alpha \in R and show that \forall \vec{x} \in S,~ \alpha \vec{x} \in S).
    So if  \vec{x}=(a_1,2a_1,3a_1) still,
    I can say that  \alpha \vec{x} = \alpha (a_1,2a_1,3a_1)
    therefore  \alpha \vec{x} = (\alpha a_1, 2 \alpha a_1, 3 \alpha a_1)

    Is that proven though?
    I didn't understand what you meant with the hint sorry.
    Thanks,
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  5. #5
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    Indeed, \alpha \vec{x} = (\alpha a, 2 (\alpha a), 3 (\alpha a))<br />
. But, we know that any vector x in S has its first component: any real, the second component: twice the first component, and the third component: three times the first. In other words, any vector \vec{v} = (a, 2a, 3a) \in S because it satisfies the conditions of membership of S.

    We can see that \alpha \vec{x} is also such a vector. To make this more clear, let \alpha a = b \in R and substitute it:

    \alpha \vec{x} = (\alpha a, 2 (\alpha a), 3 (\alpha a))=(b, 2 b, 3b). Clearly, \alpha \vec{x} is also in S.

    Because \alpha is any arbitrary real rumber, and \vec{x} is any arbitrary vector in S, we have shown that S is closed under scalar multiplication.

    Hope that makes sense.
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  6. #6
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    Ok, thanks so far.
    For the second one  (x,y,z) \mid 2x - 3y + 4z = 1

    I can see that if  x = \frac{1}{2} ,  y = 0 and  z = 0 then the the equation is valid, therefore it is not an empty set.

    if I took  \vec{a} = (x_1,y_1,z_1) and  \vec{b} = (x_2,y_2,z_2)

    Then  \vec{a} + \vec{b} = (x_1 + x_2, y_1 + y_2, z_1 + z_2)

    This still has to satisfy the condition:  2x - 3y + 4z = 1

    so I can say that  2(x_1 + x_2) - 3(y_1 + y_2) + 4(z_1 + z_2) = 1

    So is this subset of  R^3 only a subspace under certain values of  x_1, x_2, y_1, y_2, z_1, z_2 ?
    If so how do I define this?
    Thanks


    Edit:
    If I consider  \vec{a} and  \vec{b} to be on the plane  2x - 3y + 4z = 1 , then any sum of  \vec{a} and  \vec{b} will also lie on the plane. Therefore closed under vector addition. I know this isn't sufficient, but I think I may be on the right track..

    Edit #2: back to  \vec{a} = (x_1,y_1,z_1) and  \vec{b} = (x_2,y_2,z_2)
    From this,  \vec{a} = 1 and  \vec{b} = 1 therefore,  \vec{a} + \vec{b} = 2 which is not allowed as  2x - 3y + 4z = 1
    Which means that  2(x_1 + x_2) - 3(y_1 + y_2) + 4(z_1 + z_2) != 1 so therefore it is not closed under vector addition!

    Please tell me that this one is right.
    Thanks,
    Last edited by U-God; April 6th 2009 at 02:37 AM.
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  7. #7
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    Let S=\lbrace (x,y,z) \mid 2x - 3y + 4z = 1 \rbrace \subset R^3.

    The quick way to prove that it is not a subspace is by showing that \vec{0} \notin S. I.e. 2(0) - 3(0) + 4(0) \neq 1.

    But just to show that it isn't closed under vector addition:

    Let \vec{a} = (x_1,y_1,z_1), ~\vec{b} = (x_2,y_2,z_2) be any vectors in S. Therefore we know that 2x_1-3y_1+4z_1 = 1 and likewise 2x_2-3y_2+4z_2 = 1.

    Then, \vec{a}+\vec{b} = 2(x_1 + x_2) - 3(y_1 + y_2) + 4(z_1 + z_2).
    Expanding and grouping terms, we get:
    \vec{a}+\vec{b} = \underbrace{(2x_1 -3y_1 +4z_1)}_{1} + \underbrace{(2x_2 - 3y_2 + 4z_2)}_{1} = 2

    \therefore \vec{a}+\vec{b} \notin S, and as such S is not closed under addition.
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