linear function on $\mathbb{C}^{3}$ Practice Problem:
Define a linear transformation $f:\mathbb{C}^{3}\rightarrow \mathbb{C}^{3}$ by:
$$f\left ( \left ( a_{1},a_{2},a_{3} \right ) \right )=\left ( a_{1}-a_{2}+ia_{3},2a_{1}+ia_{2},\left ( 2+i \right )a_{1}-a_{3} \right )$$
Find $\mathrm{Ker}f $ and $\mathrm{Im}f$ by giving a basis for both of them.
I found the kernel to be $\left \{ a_{1}\cdot\left ( 1,2i,2+i \right ) \right \}$ where $ a_{1}\in \mathbb{C}$.
Obviously dimension of the kernel is 1 and hence dimension of the image of $f$ is 2. But how can I find the image of $f$ with its basis? Thanks
 A: For any vector spaces $\mathbf{V}$ and $\mathbf{W}$, and for any linear transformation $f\colon\mathbf{V}\to\mathbf{W}$, if $\beta$ is a basis for $\mathbf{V}$, then $f(\beta) = \{f(\mathbf{v})\mid \mathbf{v}\in\beta\}$ spans $\mathrm{Im}(f)$. (In fact, this is true if $\beta$ is a spanning set for $\mathbf{V}$; it doesn't have to be a basis, but it may as well be, since any linear dependencies that may exist among elements of $\beta$ will necessarily exist among element of $f(\beta)$ as well...)
In particular, for your $f$, taking $\beta$ to be the standard basis we have $f(1,0,0) = (1,2,2+i)$, $f(0,1,0)=(-1,i,0)$, and $f(0,0,1)=(i,0,-1)$ will necessarily span $\mathrm{Im}(f)$.
That is, $\mathrm{Im}(f) = \mathrm{span}\Bigl( (1,2,2+i), (-1,i,0), (i,0,-1)\Bigr)$.
Now, this is a spanning set for $\mathrm{Im}(f)$, not necessarily a basis for $\mathrm{Im}(f)$. But: every spanning set contains a basis. So all we have to do is go through the three vectors, discarding any vector that is a linear combination of the previous ones. Since we know (by the Rank-Nullity Theorem) that the image of $f$ has dimension $2$, noting that $(1,2,2+i)$ and $(-1,i,0)$ are linearly independent tells us that they are a basis for the image.
(Indeed, note that 
$$(i,0,-1) = \left(-\frac{2}{5}+\frac{1}{5}i\right)(1,2,2+i) + \left(-\frac{2}{5}-\frac{4}{5}i\right)(-1,i,0)$$
so the third vector in the spanning set is already in the span of the first two.)
A: A general vector of Imf is the vector you have in the definition of the transformation, that is the result of aplying f to a general vector of the domain.
You can descompose it as a linear combination of three vectors of image making : a1(1,2,2+i,)+a2(-1,i,0)+a3(i,0,-1)
And in this way you find a generator of Imf.
