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.
 A: Does $T_{M,N}(g)$ mean "the matrix of the linear transformation $g$ w.r.t the bases $M$ (domain) and $N$ (range)? If so, proceed as in the proof of the Rank Nullity Theorem.
Choose a basis of $\ker f$: you tell me $v_3=(-2,1,1)^T$ will do. Complete to a basis of $\mathbb{R}^3$ in your favourite way: I use $v_1=(0,1,0)^T$, $v_2=(0,0,1)^T$. 
Now put $w_1=Av_1$, $w_2=Av_2$, and note that as in the proof of the Rank Nullity Theorem these are linearly independent. (You can work them out and check by hand.)  Now choose $w_3$ to be linearly independent of these. 
We are done.
A: 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 $w_3 = (0,0,1).$
A: I think $T_{MN}(f)$ means matrix representation with respect to basis M and N.
i.e. $T_{MN}(f) = [f]_M^N$. The question is then to find basis M and N such that
$diag(1,1,0)=[f]_M^N= [I]_B^N \cdot A \cdot   [I]_M^B$, when $B$ is the standard basis of $R^3$. 
 ( $diag(a,b,c)$ means diagonal matrix with elements a,b,c.)
Note that this implies
$ A\cdot  [I]_M^B= [I]_N^B \cdot diag(1,1,0) $.
If you compute eigen values, two of them is nonzero complex value (denote by $c_1,c_2$) and a $0$ eigen value. ( Denote corresponding eigen vectors by $x_i$'s. )
then
$ A \cdot [ x_1 \; x_2 \; x_3 ] = [x_1 \; x_2 \; x_3] \cdot diag( c_1,c_2,0) $
$ \rightarrow A \cdot [ x_1 \; x_2 \; x_3 ] = [x_1/c_1 \; x_2/c_2 \; x_3] \cdot diag( 1,1,0) $
Therefore
$M={x_1,x_2,x_3}$ and $N={x_1/c_1,x_2/c_2,x_3}$
A: It may be helpful for you to see how linear maps, bases and matrices are related. I will take the notation $T_{AB}(f)$ to be the matrix of $f:U \rightarrow V$, where $U$ has basis $B$ and $V$ has basis $A$. 
Let $E = \{e_1, e_2, e_3\}$ be the standard basis for $\mathbb{R}^3$. Then the linear map $f$ is defined by $A = T_{EE}(f) = \begin{pmatrix}1&3&-1\\-1&0&-2\\1&1&1\end{pmatrix}$ acts on the basis $E$ as follows,
\begin{align}
f(e_1) &= &1e_1 - 1e_2 + 1e_2 \\
f(e_2) &= &3e_1 + 0e_2 + 1e_3 \\
f(e_3) &= &-1e_1 -2e_2 + 1e_3 \\
\end{align}
The equations correspond to the columns of the matrix. We will now construct some new bases for $\mathbb{R}^3$ and see how $f$ acts on them. 
For example let $a_1 = 1e_1+1e_2+1e_3, a_2 = 1e_1-1e_2, a_3 = 1e_2-1e_3$, so $F = \{a_1, a_2, a_3 \}$ is a basis for $\mathbb{R}^3$. Doing the calculations gives,
\begin{align}
f(a_1) &= &3 e_1 - 1e_2 + 3e_2 \\
f(a_2) &= &-2e_1 -1e_2 + 0e_3 \\
f(a_3) &= & 4e_1 +2e_2 + 0e_3 \\
\end{align}
So the matrix $T_{EF}(f) = \begin{pmatrix}3&-2&4\\-1&-1&2\\3&0&0\end{pmatrix}$.
For your specific question, let's take $w_1 = e_1, w_2 = e_2, $. Notice that $f(w_1), f(w_2)$ are linearly independent. So let's take $v_1 = f(w_1), v_2 = f(w_2)$. Now $v_1, v_2$ can be extended to a basis for $\mathbb{R}^3$ for example let $v_3 = e_1$, then check that $v_1, v_2, v_3$ are linearly independent. Finally let $w_3 = -2e_1 + 1e_2 + 1e3$, this is an element of $\ker(f)$ so $f(w_3) = 0$. Also we can check that $w_1, w_2, w_3$ are linearly independent and so form a basis for $\mathbb{R}^3$. 
To recap, let's express $f(w_i)$ in terms of $v_i$ and we can read off the matrix of $f$ with respect to bases $N = \{w_1, w_2, w_3\}$ and $M = \{v_1, v_2, v_3 \}$
\begin{align}
f(w_1) &= 1v_1 + 0v_2 + 0v_3 \\
f(w_2) &= 0v_1 + 1v_2 + 0v_3 \\
f(w_3) &= 0v_1 + 0v_2 + 0v_3 \\
\end{align}
So $T_{MN}(f) = \begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}$
