Did I do it correctly? What's the image and the basis of the image of this linear mapping? 
$A$ is the linear mapping $f(x)= Ax,\mathbb{R} \rightarrow \mathbb{R}$
$$f\left( \begin{pmatrix} x_{1}\\  x_{2}\\  x_{3}
\end{pmatrix}\right)= \begin{pmatrix} x_{2}-x_{3}\\ 
x_{1}+3x_{2}-2x_{3}\\  x_{1}-4x_{2}+5x_{3} \end{pmatrix}$$ 
What's the image $\text{Im}(f)$? What's the basis of the image?

So far I have only calculated the image / basis of matrix but I hope / think it's actually the same for these mappings.
So I form $f$ to a matrix $\begin{pmatrix}
0 & 1  & -1\\ 
1 & 3  & -2\\ 
1 & -4 &  5
\end{pmatrix}$
Then transpose this matrix: $\begin{pmatrix}
0  &  1  &  1\\ 
1  &  3  & -4\\ 
-1 & -2  &  5
\end{pmatrix}$
Now get as many zero lines as possible using Gauss:
Take third line and add it to second line:
$\begin{pmatrix}
0  &  1  &  1\\ 
0  &  1  &  1\\ 
-1 & -2  &  5
\end{pmatrix}$
Now make second line negative by $\cdot(-1)$ and add it to first line:
$\begin{pmatrix}
0  &   0  &   0\\ 
0  &  -1  &  -1\\ 
-1 &  -2  &   5
\end{pmatrix}$
Transpose back we have $\begin{pmatrix}
 0  &   0  &  -1\\ 
 0  &  -1  &  -2\\ 
 0  &  -1  &   5
\end{pmatrix}$
Thus $\text{Im}(f)= \text{span} \left(\left\{ \begin{pmatrix}
0\\ 
-1\\ 
-1
\end{pmatrix},\begin{pmatrix}
-1\\ 
-2\\ 
5
\end{pmatrix} \right\}\right )$ and basis $B= \left\{ \begin{pmatrix}
0\\ 
-1\\ 
-1
\end{pmatrix},\begin{pmatrix}
-1\\ 
-2\\ 
5
\end{pmatrix} \right\}$

Can you please tell me if I did it correctly and if not how to do it correctly?
 A: You start with the matrix 
$$
 f =
\left[
\begin{array}{rrr}
 0 & 1 & -1 \\
 1 & 3 & -2 \\
 1 & -4 & 5 \\
\end{array}
\right]=
\left[
\begin{array}{rrr}
 f_{1} & f_{2} & f_{3}
\end{array}
\right].
$$
This looks like an academic exercise, so we look for trivial combinations of rows or columns. We see the column property
$$
  f_{1} - f_{2} = f_{3}.
$$
The image of $f$ is the combination of all independent column vectors:
$$
  \text{Im} \left( f \right) = a_{1} f_{1} + a_{2} f_{2} =
    a_{1}
\left[
\begin{array}{r}
 0 \\
 1 \\
 1 
\end{array}
\right]
+  a_{2}
\left[
\begin{array}{r}
  1 \\
  3 \\
 -4 
\end{array}
\right] 
%
= 
%
\left[
\begin{array}{r}
  a_{2} \\
  a_{1} + 3 a_{2} \\
  a_{1} - 4 a_{2} 
\end{array}
\right].
$$
The column combination formula provides the kernel
$$
\text{Ker} \left( f \right) =
%
\left[
\begin{array}{r}
   1 \\
  -1 \\
  -1
\end{array}
\right].
$$
A minimal basis is any two linearly independent vectors from 
$$
\text{span} \left\{ \,
\left[
\begin{array}{r}
  0 \\
  1 \\
 -1 
\end{array}
\right]
, \,
\left[
\begin{array}{r}
  1 \\
  3 \\
 -4 
\end{array}
\right] \,
\right\}
$$
For example $v_{1} = \left[
\begin{array}{r}
  0 \\
  1 \\
 -1 
\end{array}
\right]$, and $v_{2} = 
\left[
\begin{array}{r}
  1 \\
  3 \\
 -4 
\end{array}
\right].$
You could provide an orthogonal basis using the process of Gram and Schmidt:
$u_{1} = \left[
\begin{array}{r}
  0 \\
  1 \\
 -1 
\end{array}
\right]$, and $u_{2} = 
\left[
\begin{array}{r}
  1 \\
  3 \\
 -4 
\end{array}
\right].$
