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Thread: Subspaces

  1. #1
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    Subspaces

    Hey everyone.

    So linear algebra... I don't know what you're talking about.

    All this subject is very vague for me, we got some boxes of matrices and we are required to do with it some operations as robots without any understanding, so I have lost...

    We have already reached to subspaces, span, etc...

    So for example I have some subsets to determine whether are subspaces or not.

    I have the set of functions from Reals to Reals (f: -> R -> R) such that f(0)*f(1)=0
    Now, I don't understand how to determine of this set is empty.
    And actually I think we can't deal with this definition of the function as I can't divide by f(0) or f(1), it can be equal to 0 maybe.

    Thanks.
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  2. #2
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    Re: Subspaces

    not empty since f(x) = x is in the set

    f(x) = 1-x is also in the set

    f(0)*f(1)=0 means either f(0)=0 or f(1)=0
    Thanks from CStudent and topsquark
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  3. #3
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    Re: Subspaces

    Quote Originally Posted by Idea View Post
    not empty since f(x) = x is in the set

    f(x) = 1-x is also in the set

    f(0)*f(1)=0 means either f(0)=0 or f(1)=0
    It's exactly one of the things here that I don't understand.
    What does it mean f(x)=x for all real x? How are we sure about that?
    How do you know that there is an x such that f(x)=1-x?
    A little weird...

    And if one of them is 0, it remains the possibility that they are both 0 and that's a problem, no?

    Thanks.
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  4. #4
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    Re: Subspaces

    Quote Originally Posted by CStudent View Post
    And if one of them is 0, it remains the possibility that they are both 0 and that's a problem, no?
    If we start with a linear space $\mathcal{V}$ over a field $\mathcal{F}$.
    Then if we have a subset $\mathcal{S}\subseteq\mathcal{V}$ such that:
    1) $0\in\mathcal{S}$
    2) $(\forall\alpha\in\mathcal{F})~\&~(\forall\{X,~Y\} \in\mathcal{S})$ it follows that $\alpha X+Y\in\mathcal{S}$ then $\mathcal{S}$ is a subspace of $\mathcal{V}$
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  5. #5
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    Re: Subspaces

    Quote Originally Posted by Plato View Post
    If we start with a linear space $\mathcal{V}$ over a field $\mathcal{F}$.
    Then if we have a subset $\mathcal{S}\subseteq\mathcal{V}$ such that:
    1) $0\in\mathcal{S}$
    2) $(\forall\alpha\in\mathcal{F})~\&~(\forall\{X,~Y\} \in\mathcal{S})$ it follows that $\alpha X+Y\in\mathcal{S}$ then $\mathcal{S}$ is a subspace of $\mathcal{V}$
    My lack of understanding is deeper.
    I know generally the dry definitions but it doesn't help me understand the whole picture.

    First, what does it mean that we have some subset S in a vector space V?
    V is a vector space at all, S is some set, what is this relation between a set and a vector space?...
    And, why this subset must consists of 0 at least?

    As for the exercise in the topic, how are we sure that 0 is in the set of these functions?
    And more generally, what does it mean f(x)=x for all real x? How are we sure about that?
    How do you know that there is an x such that f(x)=1-x?

    Thank you all.
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  6. #6
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    Re: Subspaces

    Quote Originally Posted by CStudent View Post
    First, what does it mean that we have some subset S in a vector space V?
    V is a vector space at all, S is some set, what is this relation between a set and a vector space?...
    And, why this subset must consists of 0 at least?
    Every subspace of a linear space of a linear space is itself a linear space.
    Every linear space must have a zero vector.

    Quote Originally Posted by CStudent View Post
    As for the exercise in the topic, how are we sure that 0 is in the set of these functions?.
    You must be able to prove a subset of a linear space contains 0 in order to have a subspace.
    Thanks from CStudent
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  7. #7
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    Re: Subspaces

    Quote Originally Posted by Plato View Post
    Every subspace of a linear space of a linear space is itself a linear space.
    Every linear space must have a zero vector.


    You must be able to prove a subset of a linear space contains 0 in order to have a subspace.
    I asked about the relation between sets and spaces...

    How do we know that 0 is in S?

    Thanks.
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  8. #8
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    Re: Subspaces

    Quote Originally Posted by CStudent View Post
    I asked about the relation between sets and spaces...
    How do we know that 0 is in S?
    I answered this already once. Do you read replies carefully?
    You have to show it.
    Do you understand that the set of all $2\times 2$ matrices forms a linear space over the real field?
    If $\displaystyle \mathcal{S} = \left\{ {\left[ {\begin{array}{*{20}{c}} x&0\\ 0&y \end{array}} \right]:\left\{ {x,y} \right\} \subseteq \mathbb{R}} \right\}$.
    1) How do we show that $\left[{\begin{array}{*{20}{c}} 0&0\\ 0&0 \end{array}}\right]\in\mathcal{S}~?$
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  9. #9
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    Re: Subspaces

    Quote Originally Posted by Plato View Post
    I answered this already once. Do you read replies carefully?
    You have to show it.
    Do you understand that the set of all $2\times 2$ matrices forms a linear space over the real field?
    If $\displaystyle \mathcal{S} = \left\{ {\left[ {\begin{array}{*{20}{c}} x&0\\ 0&y \end{array}} \right]:\left\{ {x,y} \right\} \subseteq \mathbb{R}} \right\}$.
    1) How do we show that $\left[{\begin{array}{*{20}{c}} 0&0\\ 0&0 \end{array}}\right]\in\mathcal{S}~?$
    Because x and y are reals and 0 is in reals then x and y can be equal to 0?

    As for your question:
    Do you understand that the set of all 22 matrices forms a linear space over the real field?

    Not really.
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  10. #10
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    Re: Subspaces

    Quote Originally Posted by CStudent View Post
    Do you understand that the set of all 22 matrices forms a linear space over the real field?
    Not really.
    Do you even know what a linear space is?
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  11. #11
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    Re: Subspaces

    Quote Originally Posted by Plato View Post
    Do you even know what a linear space is?
    Yes.
    If you give me some set of vectors so the linear space is all the vectors we have if we multiply the elements of the set by the field we play on.
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