$M= \begin{pmatrix} 1 & 0 & 1\\ 2 & 1 & 1\\ -1 & 1 & -2 \end{pmatrix}$ is a real matrix and $F: \left\{\begin{matrix} \mathbb{R}^{3} \mapsto \mathbb{R}^{3} \\ x \mapsto M \cdot x \end{matrix}\right.$ is the linear mapping given by $M$
Choose two basis $A= (v_{1},v_{2},v_{3})$ and $B= (w_{1},w_{2},w_{3})$ from $\mathbb{R}^3$ such that $T^{A}_{B}(F)=\begin{pmatrix}1&0&0\\1&1&0\\0&0&0\end{pmatrix}$
I know how to do it the other way around, I mean when basis are given and I'm supposed to determine the tansformation matrix. But how to solve it this way? I found that task on the internet and I hope the things given above are even enough. I could try to write a solution but I barely have an idea :(
Note that you can choose both bases, which is rather uncommon for maps from a vector space to itself (where one usually would want to use the same basis twice). Given that you have this freedom, the only constraint that might form an obstruction to your task is the rank of the matrix (which will be the same whatever bases you use to express things on). It is not hard to see that $A$ has rank$~2$, and so has the prescribed form $T^A_B(F)$. In order to get a last column entirely zero, you must choose $v_3\in\ker(F)$, and nonzero. With this done, choose any $v_1,v_2$ to complete to a basis. Now the first column column of $T^A_B(F)$ is saying $T(v_1)=w_1+w_2$ and the second is saying $T(v_2)=w_2$. It is not hard to solve $w_1,w_2$ from this, and you can choose $w_3$ in any way to complete to a basis. For another pair of bases, just make other choices.