# 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 :-))

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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

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

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This is not enough, you must add that you work with an endomorphism in a finite dimensional vector space (or at least with an application between two vectors space of the same finite dimension). – Najib Idrissi Jul 3 '11 at 17:33
@zulon: That's one of conditions in the answer for Andrea Mori! Therefore there is something i have already understood. – Christian Ivicevic Jul 3 '11 at 17:36

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.

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