Linear Algebra dimension of vector space

I have an argument about what is the correct answer for the question below:

We have four vectors $v_1,v_2,v_3,v_4$ such that each set of three of these vectors are linearly independent over $\mathbb{C}$.

Now we given that $W = \operatorname{span}(w_1,w_2,w_3,w_4)$ with \begin{align*} w_1 &= v_1 + iv_2 - v_3\\ w_2 &= v_2+ iv_3 - v_4\\ w_3 &= iv_1 - v_4\\ w_4 &= iv_1 + v_2 + iv_3 - 2v_4 \end{align*}

The Question: What is the dimension of $W$?

The optional answers are:

• $1$
• $2$
• $3$
• $4$
• Can not be determined

My attempt: At first I just used row operations and then got two rows of independent vectors so I did knew that $3$ and $4$ isn't the answer for sure. But - we are given that the set of 3 vectors are independent as I mention before. After a little bit of playing with those vectors, I got that $$v_4=-v_1+(1-i)v_2+(1+i)v_3.$$ After placing this $v_4$ as well in the matrix I got that I cannot determined if the dimension is $1$ or $2$, so my final answer is "can not be determined".

The dimension in general cannot be determined, it can be $1$ or $2$.

First notice that $$w_3 = iw_1+w_2, \quad w_4 = iw_1+2w_2$$ so $\operatorname{span}\{w_1,w_2,w_3,w_4\} = \operatorname{span}\{w_1,w_2\}$.

1. If $\{v_1,v_2,v_3,v_4\}$ is linearly independent, then from $\alpha w_1+\beta w_2 =0$ we get $$0 = \alpha(v_1 + iv_2 - v_3)+\beta(v_2+ iv_3 - v_4) = \alpha v_1+(i\alpha+\beta)v_2+(-\alpha+i\beta)v_3-\beta v_4 \implies \alpha=\beta=0$$ so $\{w_1,w_2\}$ is linearly independent.
2. If $v_1,v_2,v_3$ are linearly independent and e.g. $v_4 = -v_1+(1-i)v_2+(1+i)v_3$ then all four sets $$\{v_1,v_2,v_3\},\{v_1,v_2,v_4\},\{v_1,v_3,v_4\},\{v_2,v_3,v_4\}$$ are all linearly independent but $$w_2 = v_2+ iv_3 - v_4 = v_1 + iv_2 - v_3 = w_1$$ so $\{w_1,w_2\}$ are clearly linearly dependent.