Find all vectors $w$ such that the set $\{u, v, w\}$ of vectors spans $\mathbb C^3$. Consider the vectors
$v = (i,i,i)$
$w = (1,2,3)$
Find all vectors $w$ such that the set $\{u, v, w\}$ of vectors spans $\mathbb C^3$
How can I solve this, I've tried for a couple of hours already, with no luck in finding the answer.
 A: We denote $w$ by $w=(r,s,t)$. The set $\{v,u,w\}$ spans $ \mathbb C^3$ $ \iff$
$ \det\begin{bmatrix}
i & i & -i \\
1 & 2 & 3 \\
r & s & t
\end{bmatrix}\ne 0$.
Show that this is equivalent to $t-r-2s \ne 0$.
A: It's the set of all vectors $w\in\mathbb{C}^3$ which are not a linear combination of $u$ and $v$, that is, the vactors which cannot be written as $(a+bi,2a+bi,3a+bi)$ for some $a$ and some $b$ in $\mathbb C$.
A: A linear way to find for the set $\{v,u,w\}$ that spans $C^3$:
$$ (w + v + u) \vec x  = \vec 0$$
by theorem 4 in L 1.4, then use Ax = 0. And by definition of linear independent in L 1.7 must only have trivial solution, $\vec x = \vec 0 $.
Btw this is the same as determinant. Then you use raw reduction to get REF.  
then you have the same as $ A \neq \vec 0$, where $A = w + v + u$. Solve this and get $t + r - 2s \neq 0 $, thus the solution by theorem 4 in L 1.4 is:
$$t \neq 2s -r$$
Which is the same solution as with determinant, which would be wrote like this, for easiest calculations:
$$ \det\begin{bmatrix}
r & i & 1 \\
s & i & 2 \\
t & i & 3
\end{bmatrix}\ne 0$$.
