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Thread: linear space and subspaces

  1. #1
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    linear space and subspaces

    I am having trouble understanding linear space and subspace. For example:
    Given that set $\displaystyle k \subseteq R^4$ and $\displaystyle k = \{(5, -5, 0, 0)\}$
    Prove that k is not a subspace of $\displaystyle R^4$.
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  2. #2
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    Quote Originally Posted by jayshizwiz View Post
    I am having trouble understanding linear space and subspace. For example:
    Given that set $\displaystyle k \subseteq R^4$ and $\displaystyle k = \{(5, -5, 0, 0)\}$
    Prove that k is not a subspace of $\displaystyle R^4$.
    You are saying that k contains the single vector v= (5, -5, 0, 0)? A subspace must be closed under addition. Is x+ x in the subspace?

    Or- a subspace must be closed under scalar multiplication. Is 5x in k?

    Or- a subspace must contain additive inverses. What is the additive inverse of x? Is it in k?

    Or- a subspace must contain the 0 vector. Is the 0 vector in k?
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  3. #3
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    I think that's where i got confused...There is only 1 vector. I was using the elements of the vector to prove that it is a subspace...

    k = (5, -5, 0, 0)

    I was saying that there is closure under addition: 5 + (-5) = 0 $\displaystyle \subseteq k$, -5 + 0 = -5 $\displaystyle \subseteq k$ and so on...

    But then please explain why my book says this is a subspace of R^4:

    W = $\displaystyle \{(\alpha, -\alpha, 0, 0) | \alpha$ is a real number}

    Does this mean that W is also only a single vector or alpha and minus alpha contain four elements each?? This is where I'm getting very confused?? The book also says that W contains k...
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    Quote Originally Posted by jayshizwiz View Post
    I think that's where i got confused...There is only 1 vector. I was using the elements of the vector to prove that it is a subspace...

    k = (5, -5, 0, 0)

    I was saying that there is closure under addition: 5 + (-5) = 0 $\displaystyle \subseteq k$, -5 + 0 = -5 $\displaystyle \subseteq k$ and so on...
    No, it is not a matter of the individual components. x+ x= (5, -5, 0, 0)+ (5, -5, 0, 0)= (10, -10, 0, 0) which is NOT in the set.

    5x= 5(5, -5, 0, 0)= (25, -25, 0, 0) which is NOT in the set.

    And, of course, (0, 0, 0, 0) is NOT in the set.

    But then please explain why my book says this is a subspace of R^4:

    W = $\displaystyle \{(\alpha, -\alpha, 0, 0) | \alpha$ is a real number}

    Does this mean that W is also only a single vector or alpha and minus alpha contain four elements each?? This is where I'm getting very confused?? The book also says that W contains k...
    No, the notation $\displaystyle \{(\alpha, -\alpha, 0, 0)|\alpha$ is a real number} means that every vector of that form - for every real number $\displaystyle \alpha$- is in that set. (5, -5, 0, 0) is in the set, taking $\displaystyle \alpha= 5$, but so is (10, -10, 0, 0), taking $\displaystyle \alpha= 10$, (25, -25, 0, 0), taking $\displaystyle \alpha= 25$, and so is the 0 vector, (0, 0, 0, 0), taking $\displaystyle \alpha= 0$.

    Taking $\displaystyle \alpha= x$, (x, -x, 0, 0) is in the set and taking $\displaystyle \alpha= y$ so is (y, -y, 0, 0). For any real numbers a and b, a(x, -x, 0, 0)+ b(y, -y, 0, 0)= (ax+ by, -ax- by, 0, 0)= (ax+by, -(ax+by), 0, 0) which is in the set taking $\displaystyle \alpha= ax+by$.
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  5. #5
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    Thank you so much!
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