Show that there is one linear mapping $\varphi : \mathbb{Q}^3 \rightarrow \mathbb{Q}^3$ $v_{1} = \left(
\begin{array}{c}
1\\
1\\
1\\
\end{array}
\right)$ ,
$v_{2} = \left(
\begin{array}{c}
2\\
1\\
5\\
\end{array}
\right)$,
$v_{3} = \left(
\begin{array}{c}
-1\\
1\\
-1\\
\end{array}
\right)$
where $v_{1},v_{2},v_{3} \in\mathbb{Q}^3$
Show that there is exactly one linear mapping $\varphi : \mathbb{Q}^3 \rightarrow \mathbb{Q}^3$
$\varphi(v_{1}) = \left(
\begin{array}{c}
1\\
1\\
-1\\
\end{array}
\right)$,
$\varphi(v_{2}) = \left(
\begin{array}{c}
2\\
1\\
-5\\
\end{array}
\right)$,
$\varphi(v_{3}) = \left(
\begin{array}{c}
0\\
0\\
0\\
\end{array}
\right)$
Is $ \varphi $ an isomorphism?
I have checked out that $ v_{1},v_{2},v_{3}$ are linearly independent but I have no idea how to continue it. I would appreciate any kind of help.
 A: As $\{v_{1},v_{2},v_{3}\}$ is a linearly independent set of $3$ vectors , they form a basis for $\mathbb{Q}^{3}$.
So you only have to check whether $\{\phi(v_{1}),\phi(v_{2}),\phi(v_{3})\}$ form a basis for $\mathbb{Q^{3}}$ in order to figure out if it's an isomorphism or not. If so then you will have a function mapping basis to basis which will extend to a linear map which will be a bijection and hence an isomorphism.
In this case $\phi(v_{3})=0$ . So you do not have an isomorphism.
However since any vector $v=(x,y,z)$ in $\mathbb{Q^{3}}$ can be written as
$v=c_{1}v_{1}+c_{2}v_{2}+c_{3}v_{3}$.
If you define $\phi(v)=\sum_{i=1}^{3}c_{i}\phi(v_{i})$ you will get a unique linear map $\phi$.
To express $\phi$ as an explicit function, try and find a solution for variables $c_{1},c_{2},c_{3}$ such that
$(x,y,z)=(c_{1}+2c_{2}-c_{3},c_{1}+c_{2}+c_{3},c_{1}+5c_{2}-c_{3})$ (this is the only part that requires a bit of labour. it is not tough).
Then you will be able to write
$(x,y,z)=c_{1}v_{1}+c_{2}v_{2}+c_{3}v_{3}$
And finally apply $\phi$ on both sides to get the explicit formula for the transformation.
$\phi(x,y,z)=c_{1}(1,1,-1)+c_{2}(2,1,-5)$
