find the bases for the range of a linear operator and the null space  I need to find the bases for a linear operator.
Here is the question given to me:
Consider $V=\mathbb{C}_{1\times 2}$  as a vector space over the real numbers. 
let the linear operator $\tau : V \rightarrow V $ be defined by $\tau (z_{1},z_{2})=(z_{1}-\overline{z_{1}}, z_{2}-\overline{z_{2}})$
Find bases for $\tau(V)$ and $NS(V)$ and determine whether $V=\tau(V) \bigoplus NS(\tau)$
 A: First: linear operators don't have bases. Vector spaces (and subspaces) have bases. You'll note that the question does not ask for a basis for the linear operator, it asks for a basis of the range of $\tau$, and for a basis of the nullspace of $\tau$; and it so happens that both of those are vector spaces, so we can talk about bases for them.
Now, $V$ is $4$-dimensional as a vector space over $\mathbb{R}$; a possible basis is $\{(1,0), (i,0), (0,1), (0,i)\}$. 
You should also check and make sure that $\tau$ is a linear transformation if you haven't done so.
Let's consider first the nullspace of $\tau$ (it is simpler than the range). When is $(z_1,z_2)$ in the nullspace of $\tau$?
$(z_1,z_2)\in\mathbf{N}(\tau)$ if and only if $\tau(z_1,z_2)=(0,0)$, if and only if $z_1-\overline{z_1}=0$ and $z_2-\overline{z_2}=0$, if and only if $z_1=\overline{z_1}$ and $z_2=\overline{z_2}$, if and only if $z_1$ is real and $z_2$ is real. That is, the nullspace is the span of $(1,0)$ and $(0,1)$, and they give you a basis.
What is the range of $\tau$? Note that if $z = a+bi$ is a complex number, then $z-\overline{z} = (a+bi) - (a-bi) = 2bi$. So another way of writing $\tau$ is:
$$\tau(z_1,z_2) = \Bigl( 2i\mathrm{Im}(z_1), 2i\mathrm{Im}(z_2)\Bigr).$$
So you should notice that the image of $\tau$ is contained in the span of $(i,0)$ and $(0,i)$. 
Is the image of $\tau$ exactly equal to $\mathrm{span}((i,0),(0,i))$? 
Given the answer to the latter question, can you find the answer to your question of deciding whether $V = \tau(V)\oplus \mathbf{N}(\tau)$?
