Isomorphism of linear map Suppose we have an equation of  linear transformation  $\varphi : \mathbb{R}^4 \rightarrow \mathbb{R}^4$.
How to show that such transformation is isomorphic and how to find inverse isomorphism ($\varphi^{-1}$)?
EDIT: The transformation is:
$\varphi ([x_1,x_2,x_3,x_4])=[x_1+x_2+x_3, x_1+x_3+x_4,x_1+x_4,x_1+x_2+x_3+x_4]$
 A: Recall the definition of a linear isomorphism of vector spaces. In this case is sufficient for $\dim(\ker(\varphi)) = 0$ and $\dim(\mathrm{im}(\varphi)) = 4$. To help you further requires knowledge of $\varphi$.
EDIT
Ok now try to show that 
(a) if $\varphi(\mathbb{x}) = 0$ then necessarily $\mathbb{x} = 0$. This shows $\dim(\ker(\varphi)) = 0$, thus injectivity.
(b) there are four vectors whose images are linearly independent. This shows that $\dim(\mathrm{im}(\varphi)) = 4$, or surjectivity.
I suggest that trying to find a matrix representation of $\varphi$ would be a very good idea for showing both these points.
For $\varphi^{-1}$, once you have a matrix you can compute its inverse.
A: $\phi : \mathbb R^4 \to \mathbb R^4$ is given by $\phi ((x_1,x_2,x_3,x_4))=(x_1+x_2+x_3, x_1+x_3+x_4,x_1+x_4,x_1+x_2+x_3+x_4).$ Let $e_1, e_2, e_3, e_4$ be the standard basis of $\mathbb R^4.$ Then $\phi (e_1)= (1, 1, 1, 1), \phi (e_2) = (1, 0, 0, 1), \phi(e_3)= (1, 1, 0,1), \phi(e_4) = (0, 1, 1, 1).$ Now you need to show that $\{\phi(e_1), \phi(e_2), \phi(e_3), \phi(e_4) \}$ forms a basis of $\mathbb R^4.$
Using Rank-Nullity theorem, to show that a given linear transformation is an isomorphism it is sufficient to check that it maps a basis to a basis.
