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Thread: 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

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

    So, does this mean that $\displaystyle a_{2}$ and $\displaystyle a_{5}$ 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 $\displaystyle a_{1}=1$, this gave me $\displaystyle 1=a_{3}+a_{4}$. Then choosing $\displaystyle a_{3}=0$ or $\displaystyle a_{3}=1$, I obtain $\displaystyle a_{4}=1$ or $\displaystyle a_{4}=0$, respectively.

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

    Repeating this process for a different $\displaystyle a$ 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; Jun 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

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

    So, does this mean that $\displaystyle a_{2}$ and $\displaystyle a_{5}$ 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 $\displaystyle a_{1}=1$, this gave me $\displaystyle 1=a_{3}+a_{4}$. Then choosing $\displaystyle a_{3}=0$ or $\displaystyle a_{3}=1$, I obtain $\displaystyle a_{4}=1$ or $\displaystyle a_{4}=0$, respectively.

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

    Repeating this process for a different $\displaystyle a$ 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 $\displaystyle a_1-a_3-a_4=0$ you get $\displaystyle a_1=a_3+a_4.$ so an element of $\displaystyle W$ is in the form:

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

    the set $\displaystyle \{e_1,e_2,e_3,e_4 \}$ is a basis for $\displaystyle W.$
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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

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

    So, does this mean that $\displaystyle a_{2}$ and $\displaystyle a_{5}$ 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 $\displaystyle a_{1}=1$, this gave me $\displaystyle 1=a_{3}+a_{4}$. Then choosing $\displaystyle a_{3}=0$ or $\displaystyle a_{3}=1$, I obtain $\displaystyle a_{4}=1$ or $\displaystyle a_{4}=0$, respectively.

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

    Repeating this process for a different $\displaystyle a$ 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 $\displaystyle R^5$ must have 5 vectors but this is a subspace, not all of $\displaystyle R^5$. 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, $\displaystyle a_2$ and $\displaystyle a_5$ 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 $\displaystyle a_1$. You could, of course, solve for either $\displaystyle a_3$ or $\displaystyle a_4$ as functions of the other two.

    For example, solving for $\displaystyle a_3= a_1+ a_4$, we can see that if $\displaystyle a_1= 1$ and $\displaystyle a_4= 0$, then $\displaystyle a_3= 1+ 0= 1$ so (1, 0, 1, 0, 0) is another basis vector. If $\displaystyle a_1= 0$ and $\displaystyle a_4= 0$, then $\displaystyle a_3= 0+ 1= 1$ 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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