i am having trouble with one of the homework question regarding to linear algebra(vector and span)

Question

$V$ is a vector space of some dimension, with $\vec u,\vec v,\vec w$ independent set of vectors in $V$. define the subspace of $V$ given by $W = \operatorname{span}(\vec u-\vec v+\vec w, 2\vec u+\vec v, -\vec u-2\vec v+\vec w)$

the question is:

• 1) does the vector $\vec u+\vec v$ inside $\operatorname{span}(\vec u-\vec v+\vec w, 2\vec u+\vec v, -\vec u-2\vec v+\vec w)$?
• 2) consider an arbitrary vector $a\vec u+b\vec v+c\vec w$ inside $\operatorname{span}(\vec u,\vec v,\vec w)$. Find a linear equation in the variables $a,b,c$ that is true when $a\vec u+b\vec v+c\vec w$ inside $W$

What I did for the first part is that I row reduced the matrix and get the dimension of two, so that $W = \operatorname{span}(\vec u-\vec v+\vec w, 2\vec u+\vec v)$. What should I do next? Is there anything wrong so far? Can somebody please help me!!? Thank you!

Answer 1

You seem to have found that $$2u+v=(u-v+w)-(-u-2v+w)$$ which implies that $$W=\mathrm{span}(u-v+w, 2u+v).$$

We still need to:

• Determine if $u+v$ is in this span, i.e., if $$u+v=s(u-v+w)+t(2u+v)$$ for scalars $s,t$. There's a shortcut here: since $u+v$ and $2u+v$ have no $w$ components, the value of $s$ is immediate, after which the answer is immediate.

• Give necessary and sufficient conditions for $au+bv+cw \in W$. Since $u-v+w$ and $2u+v$ are linearly independent (follows from inspection, knowing that $u,v,w$ are linearly independent), we know the vectors in $W$ all have the form $$s(u-v+w)+t(2u+v).$$ Now we just rearrange this so it's of the form $au+bv+cw$. This will give $a,b,c$ in terms of $s,t$.

Answer 2

Hint. You need to determine whether or not there is a solution for $$\def\\#1{{\bf#1}} \lambda_1(\\u-\\v+\\w)+\lambda_2(2\\u+\\v)+\lambda_3(-\\u-2\\v+\\w) =\\u+\\v\ .$$ Since $\\u,\\v,\\w$ are independent, we can equate coefficients: $$\lambda_1+2\lambda_2-\lambda_3=1\ ,\quad -\lambda_1+\lambda_2-2\lambda_3=1 \ ,\quad \lambda_1+\lambda_3=0\ .$$ If you solve these you will find that there is no solution.

For part 2, do exactly the same but with $a\\u+b\\v+c\\w$ on the RHS instead of $\\u+\\v$. The vector $a\\u+b\\v+c\\w$ is in ${\rm span}(W)$ if and only if the system has a solution; you will find that this is true for certain values of $a,b,c$ and not for others.

Good luck!