If the dimension of the space is 2, that's the max number of linearly independent vectors you can have by definition of dim.
EDIT: a more direct way of looking at it:
By def of span, all vectors in S are linear combinations of u and v,including w
Can someone show me how to do this question, ?
suppose { u,v,w} is linearly dependant than , w can be written as a linear combination of u and v, for some scalars c1, ,c2,
c1u + c2v .....kw = 0
w = (-c1/k)u - (c2/k)v
i am not sure where to go from here.
If the dimension of the space is 2, that's the max number of linearly independent vectors you can have by definition of dim.
EDIT: a more direct way of looking at it:
By def of span, all vectors in S are linear combinations of u and v,including w
You can't prove this- it isn't true. Take u= <1, 0, 0>, v= <0, 0, 0> and w= <0, 1, 0> then w cannot be written as a linear combination of u and v. Another counter example would be u= <1, 0, 0>, v= <3, 0,0>, and w= <0, 1, 0>.
But that is NOT WHAT you want to prove. You want to include "if {u, v} is linearly independent".
If u, v, w are dependent there exist scalars, a, b, c, not all 0, such that au+ bv+ cw= 0. IF c= 0, we would have
au+ bv= 0 with either a or b or both non-zero which contradicts the fact that {a, b} is linearly independent. Therefore c is non-zero and then we can solve for w: w= -(au+ bv)/c= -(a/c)u- (b/c)v.
w = (-c1/k)u - (c2/k)v
i am not sure where to go from here.
I assumed you were trying to prove thumbnail, which I thought said, if u,v span S, and w belongs to S, u,v,w are linearly dependent, which is true.
But it said, linearly INdepent, which makes the proposition false, ie, can’t be proved.
Given:
$\{u,v\}$ is linearly independent.
$w \not\in \text{span}(\{u,v\})$.
To prove:
$\{u,v,w\}$ is linearly independent.
Suppose that $\alpha u + \beta v + \gamma w = 0$.
If $\gamma \neq 0$, we have:
$w = \dfrac{-\alpha}{\gamma}u + \dfrac{-\beta}{\gamma}v$, contradicting that $w$ is not in the span of $u$ and $v$.
Hence $\gamma = 0$, and thus $\alpha u + \beta v = 0$, whence by the linear independence of $\{u,v\}$, we have $\alpha = \beta = 0$, which shows that $\{u,v,w\}$ is linearly independent.
It is always a good idea in these sorts of problems to prove what you are told to prove, and not something else which you think is the same.
Step 1): Identify the question. Assume it’s the thumbnail (I can’t remember them, so):
“u, v, and w are elements of a vector space V.
{u,v} is a linearly independent set, and w ɇ span {u,v}.
Prove that {u,v,w} is an independent set.”
Step 2): The solution.
Assume {u,v,w} is dependent. Implies w ϵ span {u,v}. Contradiction.