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Thread: Question regarding spanning.

  1. #1
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    Help proving a group of vectors span R^4

    Given $\displaystyle u_1 , u_2 , u_3 , u_4$ is a basis for $\displaystyle R^4$.
    $\displaystyle v_1=ku_1-u_3+u_4$ , $\displaystyle v_2=u_1+u_2-u_4$ , $\displaystyle v_3=4u_2+ku_3-6u_4$

    a) For which values of k are vectors $\displaystyle v_1, v_2, v_3$ 1) independent 2) dependent

    b) In cases where $\displaystyle v_1, v_2, v_3$ are dependent, write $\displaystyle v_3$ as a linear combination of $\displaystyle v_1$ and $\displaystyle v_2$

    c) For which values of k does the group $\displaystyle u_1, u_2, u_3, v_1, v_2$ span $\displaystyle R^4$

    a) I proved that when k does not equal 2, they are independent. And when k = 2 they are dependent.

    b) i proved that $\displaystyle -2v_1+4v_2=v_3$

    c) and this is where i am having trouble. I came to the conclusion that the group spans $\displaystyle R^4$ as long as $\displaystyle v_1$ and $\displaystyle v_2$ both are not linear combination of $\displaystyle u_1, u_2, u_3$. (It doesn't matter if one of the two is a linear combination since there are 5 vectors) It is given that $\displaystyle u_1 , u_2 , u_3 , u_4$ is a basis for $\displaystyle R^4$. That means that $\displaystyle u_1 , u_2 , u_3$ are independent of each other. By doing the work, I got that $\displaystyle v_1$ and $\displaystyle v_2$ are both not linear combinations of $\displaystyle u_1 , u_2 , u_3$ which means it doesn't matter what value $\displaystyle k$ is. Can someone please help me with this. Thank you very much
    Last edited by jayshizwiz; Apr 1st 2010 at 05:22 AM. Reason: To make urgent question easier for users to solve
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  2. #2
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    Urgent span question. nearly complete

    I will show my work and hopefully someone will be able to help me with the last part.

    a) If $\displaystyle v_1, v_2, v_3$ are independent that means that $\displaystyle \lambda_1 v_1+\lambda_2 v_2+\lambda_3 v_3=0$. Now replace the $\displaystyle v$ vectors with their values and you get: $\displaystyle \lambda_1 (ku_1-u_3+u_4)+\lambda_2 (u_1+u_2-u_4)+\lambda_3 (4u_2+ku_3-6u_4)=0 $. Now open parenthesis and then have the $\displaystyle u$ vectors on the outside: $\displaystyle (\lambda_1 k+\lambda_2)u_1+(\lambda_2+4 \lambda_3)u_2+(-\lambda_1+\lambda_3 k)u_3+(\lambda_1-\lambda_2-6\lambda_3)u_4=0 $. It is given that $\displaystyle u_1 , u_2 , u_3 , u_4$ is a basis for $\displaystyle R^4$. That means that $\displaystyle u_1 , u_2 , u_3 , u_4$ are all independent vectors. Since we know they are independent, we also know that the numbers inside the parenthesis = 0

    $\displaystyle \lambda_1 k+\lambda_2 =0$
    $\displaystyle \lambda_2 +4\lambda_3 =0$
    $\displaystyle -\lambda_1+ \lambda_3 k =0$
    $\displaystyle \lambda_1- \lambda_2 -6\lambda_3 =0$

    and the matrix look as follows:

    $\displaystyle \left(\begin{array}{ccc} k& 1& 0\\ 0& 1& 4\\-1& 0& k\\ 0&-1&-6\end{array}\right)$

    Then we get to this matrix:

    $\displaystyle \left(\begin{array}{ccc}1&0&\frac{-4}{k}\\0&1&4\\0&0&k-\frac{4}{k}\\0&0&\frac{4k+4}{k}-6\end{array}\right)$

    Now, the only way the vectors can be dependent is if it is possible to have 2 rows of zero vectors. In a case where $\displaystyle k=2$ both the bottom rows equal 0!

    So, vectors $\displaystyle v_1, v_2, v_3$ are independent when k does not equal 2

    vectors $\displaystyle v_1, v_2, v_3$ are dependent when $\displaystyle k=2$

    b) When $\displaystyle k=2$ $\displaystyle v_1, v_2, v_3$ are dependent:

    $\displaystyle \lambda_1 v_1+\lambda_2 v_2=v_3$ Now, like before I replace the v vectors with their u vector values: $\displaystyle \lambda_1 (2u_1-u_3+u_4)+\lambda_2 (u_1+u_2-u_4)=(4u_2+2u_3-6u_4)$ and open parenthesis and then put the u vectors on the outside: $\displaystyle (2\lambda_1+\lambda_2)u_1+(\lambda_2-4)u_2+(-\lambda_1 -2)u_3+(\lambda_1 -\lambda_2 +6)u_4=0$ And just like before, since we know that $\displaystyle u_1 , u_2 , u_3 , u_4$ are independent:

    $\displaystyle 2\lambda_1+\lambda_2=0$
    $\displaystyle \lambda_2-4=0$
    $\displaystyle -\lambda_1 -2=0$
    $\displaystyle \lambda_1 -\lambda_2 +6=0$

    and from this we get that $\displaystyle \lambda_1=-2$ and $\displaystyle \lambda_2=4$ and that means:

    $\displaystyle -2v_1+4v_2=v_3$

    c) This is where I am unsure. The task is to find the value of k which would make $\displaystyle u_1, u_2, u_3, v_1, v_2$ span $\displaystyle R^4$. The rule is that you need at least 4 independent vectors to span $\displaystyle R^4$. The group they gave us $\displaystyle u_1, u_2, u_3, v_1, v_2$ contains 5 vectors and we know the the u vectors are independent. So now I need to prove that at least one of the v vectors are independent.

    So let's take $\displaystyle v_1$. If $\displaystyle v_1$ is not a linear combination of the group then we know that the group spans $\displaystyle R^4$. But let's assume it does and check: $\displaystyle \lambda_1 u_1+\lambda_2 u_2+\lambda_3 u_3=v_2$ Now replace the v vectors with their u vector values... : $\displaystyle \lambda_1 u_1+\lambda_2 u_2+\lambda_3 u_3=ku_1 -u_3+u_4$ and do what we did before to get: $\displaystyle (\lambda_1 -k)u_1+(\lambda_2)u_2+(\lambda_3 -1)u_3+(-1)u_4=0$ and since we know that $\displaystyle u_1 , u_2 , u_3 , u_4$
    is independent... :

    $\displaystyle \lambda_1 -k=0$
    $\displaystyle \lambda_2=0$
    $\displaystyle \lambda_3 -1=0$
    $\displaystyle -1=0$

    and since -1 does not equal 0 that must mean that $\displaystyle v_1$ is not a linear combination of the group which means that the group spans $\displaystyle R^4$ for ANY VALUE OF K. Can someone please tell me if I came to the write conclusion?? This is pretty urgent. I have an assignment do on Friday so I do not have a lot of time.

    Thanx so much.
    Last edited by jayshizwiz; Mar 31st 2010 at 10:31 AM.
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  3. #3
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    I feel stupid for asking this but...How do I delete a post? I am trying to combine my posts to make it neater, more understandable, and easier for people to answer my question.
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