Proof that a linear transformation is isomorphic
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 :-))
$\endgroup$ 92 Answers
$\begingroup$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.
$\endgroup$ 2 $\begingroup$By the definition the linear isomorphism is a linear bijective map. We should show that for any point $(x,y)\in \mathbb R^2$ there exists unique point $(x',y')\in \mathbb R^2$ such that $f(x',y') = (x,y)$ (this point is called pre-image of $(x,y)$).
Note that these any + unique are properties of a bijection.
The simpliest way here is to provide explicitly such a pre-image. By the construction of $f$ we know that $$(x,y) = (x'-y',x'+y').$$ On the other hand, we can extract $x'$ and $y'$ through $x$ and $y$ from the last equation. We obtain $$ \begin{cases} x'-y' = x, \\ x'+y' = y \end{cases} $$ and then $$ x' = \frac{x+y}{2},\quad y' = \frac{y-x}{2}. $$ This formulas give a unique result for any $x$ and $y$, hence we proved that $f$ is isomorphic.
$\endgroup$
