Given an isomorphism linear transformation $T:V \to \mathbb C^3$,where $V$ is a vector space over $\mathbb C$,assume $\alpha_1,\alpha_2,\alpha_3,\alpha_4 \in V$ such that: $$T(\alpha_1)=(1,0,i)$$ $$T(\alpha_2)=(-2,1+i,0)$$ $$T(\alpha_3)=(-1,1,1)$$ $$T(\alpha_4)=(\sqrt 2,i,3)$$
- Is $\alpha_1$ in the subspace spanned by $\alpha_2,\alpha_3$?
- If $W_1$ is the subspace spanned by $\alpha_1,\alpha_2$ and $W_2$ is the subspace spanned by $\alpha_3,\alpha_4$,then find the intersection of $W_1$ and $W_2$.
- Find a basis for the subspace spanned by $\alpha_1,\alpha_2,\alpha_3,\alpha_4$.
$\alpha_1$ is the subspace spanned by $\alpha_2,\alpha_3$ if it can be written as a (finite) linear combination of $\alpha_2,\alpha_3$,if there exist $\lambda_2,\lambda_3 \in \mathbb C$ such that $$\alpha_1=\lambda_2\alpha_2+\lambda_3\alpha_3$$
Or equivalently (the equivalence is duo to the fact that $T$ is an isomorphism which follows that it does have an inverse) $$T(\alpha_1)=\lambda_2T(\alpha_2)+\lambda_3T(\alpha_3)$$ $$(1,0,i)=\lambda_2(-2,1+i,0)+\lambda_3(-1,1,1)$$ Which implies that $\lambda_2=i(i+1)/2,\lambda_3=i$,and so the answer to the question is yes.
So far we know that $\text{dim}(W_1)=2=\text{dim}(W_2)$,on the other hand $$ \text{dim}(W_1 \cap W_2)=\text{dim}(W_1)+\text{dim}(W_2)- \text{dim}(W_1+W_2)$$
$T$ is isomorphism ,hence a bijection,so does have an inverse ,but I don't know how to find the dimension of $W_1+W_2$.
Since the $\alpha_i$'s have not been given explicitly,hence I think we need a trick to find such a basis,but I don't know how.