Proof that a linear transformation is isomorphic

As a 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 :-))

• Are you implying that $f(1,0)=(1,1)$ and $f(0,1)=(-1,1)$? (Note: adopting the row vector conventions saves \TeX work....) – Andrea Mori Jul 3 '11 at 16:52
• I am sorry, but i do not understand what you want to express with that (trivial) implication. – Christian Ivicevic Jul 3 '11 at 16:55
• Those are two basis elements in your domain... They could map to basis elements in your codomain... – mathmath8128 Jul 3 '11 at 16:57
• @Christian: A linear tranformation $T$ is completely detrmined by what it does to a basis. The standard technique is to arrange the images of the basis vectors in a rectangular array, called the associated matrix $M$, and infer properties of $T$ from properties of $M$ which are generally simpler to obtain. This lies at the very foundation of linear algebra and is explained in every textbook. I don't think is a good idea to attempt homework without some theoretical background – Andrea Mori Jul 3 '11 at 17:07
• @Andrea Mori: I must confess that i have some problems to understand the current contents of our lecture, but give me a try. For every linear transformation $T$ from one space $X$ to anothe one called $Y$ with dim$X$=dim$Y$ i can say that $T$ is isomorphic, when i proove bijection. Let $T$ be injective and therefore the kernel is trivial. Due to the rank–nullity theorem we have dim im T = dim X - dim ker T = dim Y and $T$ is bijective if dim im T = dim Y, that means im $T = Y$. However $T$ is surjective if im $T = Y$ what has been proven. -> Therefore $T$ is isomorphic. / Is that ok? – Christian Ivicevic Jul 3 '11 at 17:21

Clearly, $ker T=\{(x,y):T(x,y)=(0,0)\}=\{(x,y):(x-y,x+y)=(0,0)\}=\{(x,y):x-y=0,x+y=0\}$. As $x-y = 0,x+y=0$ easily imply that $x=y=0$ so $ker T=\{(0,0)\}$ and we are done.
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 $(x,y)$).
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.