TITLE: Proof that a linear transformation is isomorphic
QUESTION [5 upvotes]: As homework, I need to proove whether a few linear transformations are isomorphic or not, however i do not know how to achieve this. First of all i have proven that the following map is linear:
$$f:\mathbb{R}^2\mapsto\mathbb{R}^2, f\left( \begin{bmatrix}x\\y\end{bmatrix} \right)=x\begin{bmatrix}1\\1\end{bmatrix} + y \begin{bmatrix}-1\\1\end{bmatrix}$$
via the definition that a linear transformation $T: X\mapsto Y$ must satisfy the following condition (let $X$ and $Y$ be linear spaces over the field $\mathbb{K}$)
$$\forall\alpha,\beta\in\mathbb{K}\wedge x_1,x_2\in X\,:\,T(\alpha x_1+\beta x_2) = \alpha T x_1+\beta T x_2$$
Can you explain me where to go ahead? (I am from germany so please make your explanations not that difficult :-))

REPLY [3 votes]: See the kernel of the linear transformation. If it contains only the zero vector, then the linear transformation is one to one and hence an isomorphism (as the two spaces in question are of the same finite dimension).
Clearly,
$$\begin{aligned}
\operatorname{ker}T&=\{(x,y):T(x,y)=(0,0)\} \\
   &=\{(x,y):(x-y,x+y)=(0,0)\}\\
   &=\{(x,y):x-y=0,x+y=0\}.\end{aligned}$$
As $x-y = 0,x+y=0$ easily imply that $x=y=0$ so $\operatorname{ker} T=\{(0,0)\}$ and we are done.