Find 2 basis $M=\left\{v_1,v_2,v_3\right\}$ and $N=\left\{w_1,w_2,w_3\right\}$ such that $T_{MN(f)}$ 
$A=\begin{pmatrix} 1 & 3 & -1\\ 
-1 & 0 & -2\\  1 & 1 & 1 \end{pmatrix}$ is a real matrix and $f: \mathbb{R}^3 \rightarrow \mathbb{R}^3, \text{ } f(x)= A \cdot x$ is a
  linear mapping.
Find two basis $M= \left\{v_1,v_2,v_3\right\}$ and $N=
\left\{w_1,w_2,w_3\right\}$ of $\mathbb{R}^3$ such that $T_{MN}(f) =
\begin{pmatrix} 1 & 0 & 0\\  0 & 1 & 0\\  0 & 0 & 0 \end{pmatrix}$

I have absolutely no idea how to do this task, nor understand the notation $T_{MN}$. Maybe I could do it if I  would understand the notation. I hope some will explain me what it means / how this could be solved?
If it matters, $Ker(f)=\left\{\begin{pmatrix}
-2z\\ 
z\\ 
z
\end{pmatrix} \mid z \in \mathbb{R}\right\}$ and $Im(f)= span\left(\left\{\begin{pmatrix}
1\\ 
-1\\ 
1
\end{pmatrix},\begin{pmatrix}
0\\ 
-3\\ 
2
\end{pmatrix}\right\}\right)$
(This is no homework!)
I will set a bounty on this question of 200 rep because I really want understand it.

If $V$ is a vector space with finite dimension, then, there exists a number $d$ (and only one) such that every basis of $V$ has exactly $d$ vector; said $n$ is the dimension of $V.$
Assume $B = (v_1, \ldots, v_d)$ is one such basis of $V.$ I write a $d$-tuple instead of the more (erroneously used) common usage of a set because the order of the vectors matters. One can create a mapping, which happens to be both linear and bijective, from $V$ to $\Bbb R^d$ via the base $B$ by means of
$$v = \sum_{k = 1}^d \lambda_k v_k \mapsto (\lambda_1, \ldots, \lambda_d).$$
This mapping is called the coordinates of $V$ (relative to the chose of base $B$). Denote it by $[ \cdot ]_B:v \mapsto [v]_B.$
The coordinates, as I said before, is a linear bijective mapping and hence, it is an isomorphisim of vector spaces.
Given two finite dimensional vector spaces $V$ and $W$ with corresponding basis $M$ and $N,$ one can create the coordinate spaces $[V]_M = \Bbb R^p$ and $[W]_N = \Bbb R^q,$ where $p$ and $q$ are the respective dimensions of $V$ and $W.$ Basically, what this does is changing the "abstract" spaces $V$ and $W$ for two more familiar ones, viz $\Bbb R^p$ and $\Bbb R^q.$
For a linear transformation $f:V \to W$ one can create a matrix and only one $T_{M,N} = [f]_M^N$ with the defining property that
$$[f]_M^N[v]_M = [f(v)]_N, \quad v \in V.$$
To your exercise. You know that $f(\xi_1, \xi_2, \xi_3) = (\xi_1 + 3\xi_2 - \xi_3, -\xi_1 - 2\xi_3, \xi_1 + \xi_2 + \xi_3).$ So, you can calculate $f(v_1), f(v_2)$ and $f(v_3).$ You want a basis $N = (w_1, w_2, w_3)$ of $\Bbb R^3$ such that $[f(v_1)]_N = (1, 0, 0),$ $[f(v_2)]_N = (0,1,0)$ and $[f(v_3)]_N = (0,0,0).$ Notice that you want $[f(v_3)]_N = 0$ (so by means of the isomorphism, $v_3 \in \ker f$) and $f(v_1)$ and $f(v_2)$ are the first and second vector in $N.$ The natural candidate for $M$ is $v_1 = (1, -1, 1), v_2 = (0, -3, 2)$ and $v_3 = (-2, 1, 1).$ Make $N = (f(v_1), f(v_2), w_3),$ where $w_3$ is any vector that completes the base. Since $f(v_1) = (3,-3,1), f(v_2) = (-11,-4,-1)$ one can take $w_1 = (3,-3,1),$ $w_2 = (-11,-4,-1)$ and $w_3 = (0,0,1).$
