Showing a basis of $V$ over $\mathbb C$ 
Given that $V$ is over $\mathbb{C}$ and $\{u, v, w\}$ are distinct and a basis of $V$. Show that $$\{u - (1 +i)v, u+v+w, -2iu\}$$ is also a basis.

So far, I have just started by stating that they must be linear combinations of $V$. Such that:
$$u = a_1 (u-(1+i)v),\ v = a_2 (u+v+w),\ w = a_3 (-2iu).$$
But I am unsure how to proceed. 
 A: Here's a direction to help get you going:
To be a basis, those three new vectors --- let's call them $u', v', w'$ --- have to be linearly independent, and have to span $V$. 
Suppose that 
$$
a_1 u' + a_2 v' + a_3 w' = 0.
$$
By replacing $u'$ with $u + (1+i)v$, and similarly for the other two, you get an equation involving $u,v, w$. 


*

*What are the coefficients in that equation?

*What can you say about the coefficients, knowing that $u,v, w$ are linearly independent? 

*What does that tell you about $a_1, a_2, a_3$? 
If all that works out, you'll have shown linear independence. What do you have to do to show that the three new vectors also span $V$? [If you know theorems about dimension, this is a lot easier. If not, you've got some actual work to do.]
A: Another, possibly faster way, to show it would be to use a transformation matrix for a change of basis but you need to know a bit of theory behind.
Imagine the new vectors as images of the old vectors in a linear map. Write the matrix of this map (in the old basis) - you get a 3x3 matrix. Show that the matrix has full rank (which is equivalent to invertibility for square matrices). As a consequence you get that the new vectors form a basis as well.
