I have two vectors: $u = (1, -1, 0, 1)$ and $v = (-2, 2, 1, 1)$. My goal is to verify whether they are independent in R3. I haven't been able to find a linear combination of the two that equals $0$, but I did notice that $u + \frac{1}{2}v = (0, 0, 1/2, 3/2)$.
My question is, does the above mean that the two are independent and form a basis of R3, or are they only independent in R2?
Frist, your vectors have $4$ coordinates so they don't belong to $\mathbb{R}^3$ so you cant' even talk them being independant in $\mathbb{R}^3$. Next, they can't form a basis of $\mathbb{R}^3$ nor $\mathbb{R}^4$ as a basis of a vector space of dimension $n$ always has $n$ vectors, and the dimension of $\mathbb{R}^n$ seen as a real vector space is $n$.
Now go back to the definition of independant vectors in a vector space. $u$ and $v$ are not independant iff there exists a pair of reals $(a,b)\neq(0,0)$ such that $au+bv=0$. As none of the vectors is $0$ this amounts to the existence of a real $a$ such that $au=v$.
Suppose such a real $a$ exists, then one has $(a,-a,0,a)=au=v=(-2,2,1,1)$. The equality of vectors means the equality component-wise so we get $a=-2$ by the first component and $a=1$ by the 4th component, which is a contradiction. Hence such a real does not exist and the two vectors are linearly independant in $\mathbb{R}^4$.