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Thread: Subspaces (basic)

  1. #1
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    Subspaces (basic)

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

    $\displaystyle a)$ $\displaystyle (a,2a,3a) \mid a \epsilon R $
    $\displaystyle b)$ $\displaystyle (x,y,z) \mid 2x - 3y + 4z = 1 $
    $\displaystyle c)$ $\displaystyle (1,a,b) \mid a,b \epsilon R $
    $\displaystyle d)$ $\displaystyle (a-2b,b,a+b) \mid a,b \epsilon R $
    $\displaystyle e)$ $\displaystyle (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 $\displaystyle a) $ is going to be a subspace of $\displaystyle R^3 $ because if $\displaystyle 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: $\displaystyle S=\lbrace (a,2a,3a) \mid a \in R \rbrace \subset R^3$.

    Closure under vector addition:

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

    Scalar multiplication can be showed similarly. (Hint, define an $\displaystyle \alpha \in R$ and show that $\displaystyle \forall \vec{x} \in S,~ \alpha \vec{x} \in S$).
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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: $\displaystyle S=\lbrace (a,2a,3a) \mid a \in R \rbrace \subset R^3$.

    Closure under vector addition:

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

    Scalar multiplication can be showed similarly. (Hint, define an $\displaystyle \alpha \in R$ and show that $\displaystyle \forall \vec{x} \in S,~ \alpha \vec{x} \in S$).
    So if $\displaystyle \vec{x}=(a_1,2a_1,3a_1) $still,
    I can say that $\displaystyle \alpha \vec{x} = \alpha (a_1,2a_1,3a_1) $
    therefore $\displaystyle \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, $\displaystyle \alpha \vec{x} = (\alpha a, 2 (\alpha a), 3 (\alpha a))
    $. 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 $\displaystyle \vec{v} = (a, 2a, 3a) \in S$ because it satisfies the conditions of membership of S.

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

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

    Because $\displaystyle \alpha$ is any arbitrary real rumber, and $\displaystyle \vec{x}$ 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 $\displaystyle (x,y,z) \mid 2x - 3y + 4z = 1 $

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

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

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

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

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

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


    Edit:
    If I consider $\displaystyle \vec{a} $ and $\displaystyle \vec{b} $ to be on the plane $\displaystyle 2x - 3y + 4z = 1 $, then any sum of $\displaystyle \vec{a} $ and $\displaystyle \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 $\displaystyle \vec{a} = (x_1,y_1,z_1) $ and $\displaystyle \vec{b} = (x_2,y_2,z_2) $
    From this, $\displaystyle \vec{a} = 1 $ and $\displaystyle \vec{b} = 1 $ therefore, $\displaystyle \vec{a} + \vec{b} = 2 $ which is not allowed as $\displaystyle 2x - 3y + 4z = 1 $
    Which means that $\displaystyle 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; Apr 6th 2009 at 02:37 AM.
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  7. #7
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    Let $\displaystyle S=\lbrace (x,y,z) \mid 2x - 3y + 4z = 1 \rbrace \subset R^3$.

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

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

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

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

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