Show that $\{u,v,w\} \neq 〈{u+v, u+v+w, u+v+ 2w}〉$

Question

Suppose that $\{u,v,w\}$ is an independent set in some vector space $V$

How would I show that $\{u,v,w\} \neq 〈{u+v, u+v+w, u+v+ 2w}〉$?

Do I have to make use of the definition of linear independence which is

We say $T \subseteq V$ is linearly independent if whenever

$c_{1}t_{1} + c_{2}t_{2} + ... + c_{n}t_{n} = \vec 0$ for any $t_{1},...,t_{n} \in T$ ,we must have $c_{1} = 0, c_{2} = 0, c_{m} = 0$

Answer 1

The set $\{u,v,w\}$ only has three elements, whereas $\langle u+v, u+v+w, u+v+ 2w\rangle$ has infinitely many, since it is a vector space which is not $\{0\}$ (assuming that you are working over an infinite field).

Answer 2

Assuming that you are trying to see that $\langle u,v,w\rangle\neq \langle u+v,u+v+w,u+v+2w\rangle$ one solution is observe that $(u+v)-2\cdot(u+v+w)-(u+v+2w)=0$ so these three vectors are not linearly independent. Therefore they cannot span a $3$-dimensional vector space.