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Math Help - Finding a basis

  1. #1
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    Finding a basis

    Can anyone verify that I am doing this correctly?

    I have to find a basis for

    W=\{(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in F^{5} | a_{1}-a_{3}-a_{4}=0\}

    So, does this mean that a_{2} and a_{5} can be any numbers?

    I began by setting them both equal to one, giving me (0, 1, 0, 0 ,0) and (0, 0, 0, 0, 1)

    Then, by letting a_{1}=1, this gave me 1=a_{3}+a_{4}. Then choosing a_{3}=0 or a_{3}=1, I obtain a_{4}=1 or a_{4}=0, respectively.

    So that gave me the vectors (1, 0, 0, 1, 0) and (1, 0, 1, 0, 0).

    Repeating this process for a different a gave me
    (0, 0, 1, 1, 0).

    That's a total of 5 vectors, which is how many the basis should have, correct? Can anyone verify this solution? Is this a correct method to use? I just feel weird "choosing" the a's to be 1 or 0 and stuff.
    Last edited by paupsers; June 1st 2009 at 09:08 PM.
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  2. #2
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    Quote Originally Posted by paupsers View Post
    Can anyone verify that I am doing this correctly?

    I have to find a basis for

    W=\{(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in F^{5} | a_{1}-a_{3}-a_{4}=0\}

    So, does this mean that a_{2} and a_{5} can be any numbers? yes!

    I began by setting them both equal to one, giving me (0, 1, 0, 0 ,0) and (0, 0, 0, 0, 1)

    Then, by letting a_{1}=1, this gave me 1=a_{3}+a_{4}. Then choosing a_{3}=0 or a_{3}=1, I obtain a_{4}=1 or a_{4}=0, respectively.

    So that gave me the vectors (1, 0, 0, 1, 0) and (1, 0, 1, 0, 0).

    Repeating this process for a different a gave me
    (0, 0, 1, 1, 0).

    That's a total of 5 vectors, which is how many the basis should have, correct? Can anyone verify this solution? Is this a correct method to use? I just feel weird "choosing" the a's to be 1 or 0 and stuff.
    no, it's not quite correct. from the relation a_1-a_3-a_4=0 you get a_1=a_3+a_4. so an element of W is in the form:

    (a_3+a_4,a_2,a_3,a_4,a_5)=a_2e_1+a_3e_2+a_4e_3+a_5  e_4, where e_1=(0,1,0,0,0), \ e_2=(1,0,1,0,0), \ e_3=(1,0,0,1,0), and e_4=(0,0,0,0,1).

    the set \{e_1,e_2,e_3,e_4 \} is a basis for W.
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  3. #3
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    Quote Originally Posted by paupsers View Post
    Can anyone verify that I am doing this correctly?

    I have to find a basis for

    W=\{(a_{1}, a_{2}, a_{3}, a_{4}, a_{5} \in F^{5} | a_{1}-a_{3}-a_{4}=0\}

    So, does this mean that a_{2} and a_{5} can be any numbers?

    I began by setting them both equal to one, giving me (0, 1, 0, 0 ,0) and (0, 0, 0, 0, 1)

    Then, by letting a_{1}=1, this gave me 1=a_{3}+a_{4}. Then choosing a_{3}=0 or a_{3}=1, I obtain a_{4}=1 or a_{4}=0, respectively.

    So that gave me the vectors (1, 0, 0, 1, 0) and (1, 0, 1, 0, 0).

    Repeating this process for a different a gave me
    (0, 0, 1, 1, 0).

    That's a total of 5 vectors, which is how many the basis should have, correct? Can anyone verify this solution? Is this a correct method to use? I just feel weird "choosing" the a's to be 1 or 0 and stuff.
    No. A basis for all of R^5 must have 5 vectors but this is a subspace, not all of R^5. Typically each equation restricting values reduces the dimension by 1. Since this space is restricted by one equation, we would expect the dimension to be 4 and expect a basis to contain four vectors.
    Yes, a_2 and a_5 have no restrictions so (0, 1, 0, 0, 0) and (0, 0, 0, 0, 1) are basis vectors for it. NonCommAlg solved the single equation for a_1. You could, of course, solve for either a_3 or a_4 as functions of the other two.

    For example, solving for a_3= a_1+ a_4, we can see that if a_1= 1 and a_4= 0, then a_3= 1+ 0= 1 so (1, 0, 1, 0, 0) is another basis vector. If a_1= 0 and a_4= 0, then a_3= 0+ 1= 1 so (0, 0, 1, 1, 0) is a fourth basis vector. Those four vectors, {(0, 1, 0, 0, 0), (0, 0, 0, 0, 1), (1, 0, 1, 0, 0), (0, 0, 1, 1, 0)} form a basis for this subspace slightly different from the one NonCommAlgebra gave.
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