linear algebra, dimension of vector of subspace Suppose that $v$, $w$ are linearly independent vectors of $R^3$ then the subspace of $R^3$ generated by the vecors $v$,$w$ and $v-w$ is of dimension 

$A)\quad 1$

$B)\quad2$

$C)\quad3$

$D)\quad-1$

I am thinking in this question as $v$, $w$ are linearly independent so $v-w$ is linearly dependent so dimension of linearly dependent vectors is always less then $\dim(R^3)$ so is the answer $2$ ? am I correct or not? Please help me out to solve this.
HINT: $w$ and $v$ are linearly independent, so dimension is at least 2. Why the dimension of a space generated by $v$, $w$, $v-w$ must be less than 3?

Your selection is correct, but your answer doesn't explain, why the dimension is not equal to 1.
Note that $v-w$ is a linear combination of $v$ and $w$ that is


$v-w=1\cdot v-1\cdot w$


or also that in the basis $\{v,w,z\}$, with $z$ a third linearly independent vector, we have


$v=(1,0,0)$
$w=(0,1,0)$
$v-w=(1,-1,0)$