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Can I show my solution? Linear Algebra

  1. Feb 18, 2004 #1
    Here's the question.

    A theorem says that any linearly independent subset S of vector space V can be extended to a basis for V.

    1. S = {(1, 1, 0, 0) ; (1, 0, 1, 0)}, V = R^4.

    In this part, the two vectors are linearly independent by verification. Then making a basis T with dimension of 4. I arrived at an answer of

    T = { (1, 1, 0, 0) ; (1, 0, 1, 0); (1, 0, 0, 0); (0, 0, 0,1)}
    Am I correct?

    2. S = { t^3 - t + 1; t^3 + 2} , V = P(sub3)

    Again, these are linearly independent by verification.
    I used the natural basis { t^3, t^2, t, 1}.

    And then, since the dimension is 4, here's my answer.

    T = {t^3 -t +1; t^3 + 2; t^3; t^2}

    But my book has a different answer....
    instead of t^3 and t^2, it was t and 1 respectively. Are both
    of these correct?
     
  2. jcsd
  3. Feb 18, 2004 #2

    HallsofIvy

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    1 You should be able to check your answer yourself. Any set of 4 vectors is a basis for R4 if and only if they are linearly independent. Are (1, 1, 0, 0) ; (1, 0, 1, 0); (1, 0, 0, 0); (0, 0, 0,1) linearly independent? If a(1, 1, 0, 0)+ b(1, 0, 1, 0)+ c(1, 0, 0, 0)+ d(0, 0, 0,1)= (0, 0, 0, 0) what are a, b, c, and d?


    2 It's alright that your book has a different answer. Given a set of independent vectors smaller than a basis, there exist an infinite number of bases containing that set.

    Again, if a(t^3 -t +1)+ b(t^3 + 2)+ c(t^3)+ d(t^2)= 0 (for all t) what are a, b, c, and d?
     
  4. Feb 18, 2004 #3
    Hello again sir.

    Sir,

    so you are implying that if the scalars a, b, c, d are all zeros, then my answer for the basis is correct?
     
  5. Feb 18, 2004 #4

    enigma

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    No, it needs to be a non-trivial solution.
     
  6. Feb 19, 2004 #5
    ...

    So, to be nontrivial is to have at least a scalar not equal to 0 right?

    I hope you don't mind helping me here... in my questions.

    1. S = {(1, 1, 0, 0) ; (1, 0, 1, 0)}, V = R^4.

    In this part, the two vectors are linearly independent by verification. Then making a basis T with dimension of 4. I arrived at an answer of

    T = { (1, 1, 0, 0) ; (1, 0, 1, 0); (1, 0, 0, 0); (0, 0, 0,1)}
    According to our professor, I must include some of the natural basis from left to right. But how am I going to show that my basis T has a nontrivial solution? ANd as well as in 2.
     
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