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Math Help - Finding basis problem

  1. #1
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    Finding basis problem

    Given S= [a,b,c,d] belongs to R4, a not = 0. Find a basis that is a subset of s.

    I am unsure what the a not = 0 means.

    Would it just be a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1) a not = 0.

    If not, can someone give me an example of the basis.

    Thanks in advance
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  2. #2
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    Quote Originally Posted by firebio View Post
    Given S= [a,b,c,d] belongs to R4, a not = 0. Find a basis that is a subset of s.

    I am unsure what the a not = 0 means.

    Would it just be a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1) a not = 0.

    If not, can someone give me an example of the basis.

    Thanks in advance
    it is clear that v_{1}=[a,0,0,0],v_{2}=[a,1,0,0],v_{3}=[a,0,1,0],v_{4}=[a,0,0,1] are subset of S.

    now, we need to prove that v_{1},v_{2},v_{3},v_{4} are linearly independent.
    let \lambda_{1}v_{1}+\lambda_{2}v_{2}+\lambda_{3}v_{3}  +\lambda_{4}v_{4}=0,
    then we get \lambda_{2}=\lambda_{3}=\lambda_{4}=0
    and a\lambda_{1} +a\lambda_{2}+a\lambda_{3}+a\lambda_{4}=0
    because \lambda_{2}=\lambda_{3}=\lambda_{4}=0, then a\lambda_{1}=0.
    because a \not= 0, then \lambda_{1}=0
    hence, v_{1},v_{2},v_{3},v_{4} are linearly independent.
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  3. #3
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    Quote Originally Posted by dedust View Post
    it is clear that v_{1}=[a,0,0,0],v_{2}=[a,1,0,0],v_{3}=[a,0,1,0],v_{4}=[a,0,0,1] are subset of S.

    now, we need to prove that v_{1},v_{2},v_{3},v_{4} are linearly independent.
    let \lambda_{1}v_{1}+\lambda_{2}v_{2}+\lambda_{3}v_{3}  +\lambda_{4}v_{4}=0,
    then we get \lambda_{2}=\lambda_{3}=\lambda_{4}=0
    and a\lambda_{1} +a\lambda_{2}+a\lambda_{3}+a\lambda_{4}=0
    because \lambda_{2}=\lambda_{3}=\lambda_{4}=0, then a\lambda_{1}=0.
    because a \not= 0, then \lambda_{1}=0
    hence, v_{1},v_{2},v_{3},v_{4} are linearly independent.
    probably you should check your vectors
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  4. #4
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    Quote Originally Posted by firebio View Post
    Given S= [a,b,c,d] belongs to R4, a not = 0. Find a basis that is a subset of s.

    I am unsure what the a not = 0 means.

    Would it just be a(1,0,0,0)+b(0,1,0,0)+c(0,0,1,0)+d(0,0,0,1) a not = 0.

    If not, can someone give me an example of the basis.

    Thanks in advance
    That simply means vectors of the form [a, b, c, d] where a is not 0. But that also means that [0, 0, 0, 0] is not in the set nor is the scalar product of 0 with any member of the set. This is NOT a subspace of R^4 and so does not have a basis!
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  5. #5
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    Quote Originally Posted by HallsofIvy View Post
    That simply means vectors of the form [a, b, c, d] where a is not 0. But that also means that [0, 0, 0, 0] is not in the set nor is the scalar product of 0 with any member of the set. This is NOT a subspace of R^4 and so does not have a basis!
    oh,..my bad,..
    i thought the problem said that S is a subset of R^4 and we need to find the basis for vector space S. I also didn't check whether it is closed under scalar multiplication or not.

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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    That simply means vectors of the form [a, b, c, d] where a is not 0. But that also means that [0, 0, 0, 0] is not in the set nor is the scalar product of 0 with any member of the set. This is NOT a subspace of R^4 and so does not have a basis!
    if a is any real,what would be the basis ?
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  7. #7
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    Quote Originally Posted by Raoh View Post
    if a is any real,what would be the basis ?
    if a is any real, then S = R^4 and any basis for R^4 is also basis for S
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    Quote Originally Posted by dedust View Post
    if a is any real, then S = R^4 and any basis for R^4 is also basis for S
    but when you answered his question you assumed that a is any real,right ?
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  9. #9
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    Quote Originally Posted by HallsofIvy View Post
    That simply means vectors of the form [a, b, c, d] where a is not 0. But that also means that [0, 0, 0, 0] is not in the set nor is the scalar product of 0 with any member of the set. This is NOT a subspace of R^4 and so does not have a basis!
    That was what i thought at first too... but i didnt think it was suppose to be the kind of question with a strange twist to it..

    And a is any real number not = to 0?
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  10. #10
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    Someone corrected my interpretation of this question. I though it was asking for a basis for the "subspace" S but in fact it is asking for a basis for R^4 consisting of vectors in the set S.

    Try this: the standard basis for R^4 is <1, 0, 0, 0>, <0, 1, 0, 0>, <0, 0, 1, 0>, and <0, 0, 0, 1> where the last three are not from set S. Okay, just change the first component: try <1, 0, 0, 0>, <1, 1, 0, 0>, <1, 0, 1, 0>, and <1, 0, 0, 1>. Is that a basis for R^4? There are four vectors in it; are they independent?
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  11. #11
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    Quote Originally Posted by HallsofIvy View Post
    Someone corrected my interpretation of this question. I though it was asking for a basis for the "subspace" S but in fact it is asking for a basis for R^4 consisting of vectors in the set S.

    Try this: the standard basis for R^4 is <1, 0, 0, 0>, <0, 1, 0, 0>, <0, 0, 1, 0>, and <0, 0, 0, 1> where the last three are not from set S. Okay, just change the first component: try <1, 0, 0, 0>, <1, 1, 0, 0>, <1, 0, 1, 0>, and <1, 0, 0, 1>. Is that a basis for R^4? There are four vectors in it; are they independent?
    i think the standard basis of R^4 is also a basis of S.
    no ?
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  12. #12
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    Quote Originally Posted by Raoh View Post
    but when you answered his question you assumed that a is any real,right ?
    any real but not 0
    Quote Originally Posted by HallsofIvy View Post
    Someone corrected my interpretation of this question. I though it was asking for a basis for the "subspace" S but in fact it is asking for a basis for R^4 consisting of vectors in the set S.

    Try this: the standard basis for R^4 is <1, 0, 0, 0>, <0, 1, 0, 0>, <0, 0, 1, 0>, and <0, 0, 0, 1> where the last three are not from set S. Okay, just change the first component: try <1, 0, 0, 0>, <1, 1, 0, 0>, <1, 0, 1, 0>, and <1, 0, 0, 1>. Is that a basis for R^4? There are four vectors in it; are they independent?
    yes,they are linearly independent, as I proved before
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