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

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

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

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

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

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

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

c) and this is where i am having trouble. I came to the conclusion that the group spans $R^4$ as long as $v_1$ and $v_2$ both are not linear combination of $u_1, u_2, u_3$. (It doesn't matter if one of the two is a linear combination since there are 5 vectors) It is given that $u_1 , u_2 , u_3 , u_4$ is a basis for $R^4$. That means that $u_1 , u_2 , u_3$ are independent of each other. By doing the work, I got that $v_1$ and $v_2$ are both not linear combinations of $u_1 , u_2 , u_3$ which means it doesn't matter what value $k$ is. Can someone please help me with this. Thank you very much

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

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

and the matrix look as follows:

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

Then we get to this matrix:

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

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 $k=2$ both the bottom rows equal 0!

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

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

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

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

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

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

$-2v_1+4v_2=v_3$

c) This is where I am unsure. The task is to find the value of k which would make $u_1, u_2, u_3, v_1, v_2$ span $R^4$. The rule is that you need at least 4 independent vectors to span $R^4$. The group they gave us $u_1, u_2, u_3, v_1, v_2$ 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 $v_1$. If $v_1$ is not a linear combination of the group then we know that the group spans $R^4$. But let's assume it does and check: $\lambda_1 u_1+\lambda_2 u_2+\lambda_3 u_3=v_2$ Now replace the v vectors with their u vector values... : $\lambda_1 u_1+\lambda_2 u_2+\lambda_3 u_3=ku_1 -u_3+u_4$ and do what we did before to get: $(\lambda_1 -k)u_1+(\lambda_2)u_2+(\lambda_3 -1)u_3+(-1)u_4=0$ and since we know that $u_1 , u_2 , u_3 , u_4$
is independent... :

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

and since -1 does not equal 0 that must mean that $v_1$ is not a linear combination of the group which means that the group spans $R^4$ 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.

3. 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.