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Math Help - Sub-space?.. :D

  1. #1
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    Sub-space?.. :D

    Is this a subspace, if so, why and how?

    1) set of all vectors from R^n which have a first coordinate of 0.

    2) set of all vectors from R^2 of form

     \begin{bmatrix}x\\y \end{bmatrix} with x - y =2

    3) set of all vectors from R^3 of form

     \begin{bmatrix}x\\2y-x\\x-y \end{bmatrix}

    4) set of all vectors  \begin{bmatrix}x1\\x2\\x3\\x4\\x5 \end{bmatrix} from R^5 satisfying x1 + 2x2 = x3 & x4=2x5
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  2. #2
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    Just check if it satisfies scalar multiplication and addition properties. Also check if the Zero vector is contained in your set(all though this is redundant.) Can you do this?
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  3. #3
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    Not really.. I havent been given an example of how to go about tackeling a question like this.. So a starting help would be appreciated
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  4. #4
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    For instance, the first example is very simple:
    Let n\geq2 (for easier notations) be an integer and A the set of vectors in \mathbb{R}^{n} which have a first coordinate of 0.

    Clearly, 0_{\mathbb{R}^{n}}=(0,...,0) belongs to A. (\star)

    Let x=(x_{1},...,x_{n}) and y=(y_{1},...,y_{n}) be two elements in A. That means x_{1}=y_{1}=0.
    Let \lambda be a real number, then
    \lambda x + y=\lambda(0,x_{2},...,x_{n})+(0,y_{2},...,y_{n})=(  \lambda .0,\lambda x_{2},...,\lambda x_{n})+(0,y_{2},...,y_{n})= (0,\lambda x_{2}+y_{2},...,\lambda x_{n}+y_{n}) \in A

    Thus every linear combination of vectors in A is in A. (\star \star)

    (A \subset \mathbb{R}^{n}), (\star) and (\star \star) \Rightarrow A is a subspace of \mathbb{R}^{n}
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  5. #5
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    This does make sense.. I think ive successfully completed questions 2 and 3.. Ive attempted question 4, but became stuck on a number of occassions, so any help there would be much appreciated!

    Many thanks
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  6. #6
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    Maybe show your attempts for 4)
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  7. #7
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    Quote Originally Posted by brd_7 View Post
    Is this a subspace, if so, why and how?
    ...
    4) set of all vectors  \begin{bmatrix}x1\\x2\\x3\\x4\\x5 \end{bmatrix} from R^5 satisfying x1 + 2x2 = x3 & x4=2x5
    \forall\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{  bmatrix}=\begin{bmatrix}x_1\\x_2\\x_1 + 2x_2\\2x_5\\x_5 \end{bmatrix}\in\mathbb{R}^{5}
    c_1*\begin{bmatrix}x_1\\x_2\\x_1+2x_2\\2x_5\\x_5\e  nd{bmatrix}=A\subset\mathbb{R}^5
    \forall c_1\in\mathbb{R}
    \Rightarrow A is a subspace of \mathbb{R}^{5}

    I'm doing it wrong.
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  8. #8
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    Quote Originally Posted by Skerven View Post
    \forall\begin{bmatrix}x_1\\x_2\\x_3\\x_4\\x_5\end{  bmatrix}=\begin{bmatrix}x_1\\x_2\\x_1 + 2x_2\\2x_5\\x_5 \end{bmatrix}\in\mathbb{R}^{5}
    c_1*\begin{bmatrix}x_1\\x_2\\x_1+2x_2\\2x_5\\x_5\e  nd{bmatrix}=A\subset\mathbb{R}^5
    \forall c_1\in\mathbb{R}
    \Rightarrow A is a subspace of \mathbb{R}^{5}

    I'm doing it wrong.
    Only because just showing "if v in S, then av in S" is not enough to show that S is a subspace. You also have to show "if u and v in S then u+ v is in S".

    Let u= \begin{bmatrix}x_1\\x_2\\x_1 + 2x_2\\2x_5\\x_5 \end{bmatrix} and let v= [tex]\begin{bmatrix}y_1\\y_2\\y_1 + 2y_2\\2y_5\\y_5 \end{bmatrix}[/quote]

    Then u+ v= \begin{bmatrix}x_1+ y_1\\x_2+ y_2\\(x_1 + 2x_2)+ (y_1+ 2y_2\\2x_5+ 2y_5\\x_5+ y_5 \end{bmatrix}
    = \begin{bmatrix}x_1+ y_1\\x_2+ y_2\\(x_1 + y_1)+  2(x_2+ y_2)\\2(x_5+ y_5)\\x_5+ y_5 \end{bmatrix}
    which is again of the same form.

    By the way, Scopur, showing that the 0 vector is in the set is not entirely redundant. If S= {}, the empty set, then it is trivial that "if v is in S, then cv is in S" and "if u and v are in S then u+ v is in S" but the standard definition of subspace requires that it be non-empty. You must show that there is some vector in the set and it is usually easiest to show that the 0 vector is in the set.
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