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Let be $\mathbb{R}^2/\langle(1,2)\rangle$ and $\mathbb{R}$ .These two vector spaces have the same dimension $(\dim=1)$, then there is a theorem that ensures that they are isomorphic, but my question is what is that isomorphism and how to find it? thank you very much, I'm new in this topic.

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There are more than one isomorphism, so if you want to find one in particular you can choose $\{(1,0),(1,2)\}$ as a basis of $\mathbb{R}^2$, then a basis for the quotient is $[(1,0)]$ and an isomorphism is given by the linear map sending $ [(1,0)]\longmapsto 1\in \mathbb{R}$, where we have choose $1$ as a basis for $\mathbb{R}$.

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  • $\begingroup$ Thank you for your answer. How would the explicit form of the linear transformation T? $T([x,y])=x$? $\endgroup$ – Daniela Rondón Jun 2 '16 at 12:42
  • $\begingroup$ No, the map is $T([x,y])=\frac{2x-y}{2}$, infact every element in the class of $[1,2]$ is sent to zero. $\endgroup$ – InsideOut Jun 2 '16 at 13:04
  • $\begingroup$ So this map is an isomorphism from the quotient space to R. $\endgroup$ – InsideOut Jun 2 '16 at 13:06
  • $\begingroup$ thanks again, but how do you do to determine T? Can you explain please? $\endgroup$ – Daniela Rondón Jun 2 '16 at 13:06
  • $\begingroup$ Is not very easy without a picture. However I suggest you to draw in the real plane the equivalent class, which are a sheaf of parallel straight line in the plane. In particular the unique line passing through the origin is associate to the class $[1,2]$. Then using some trigonometry properties you find T. $\endgroup$ – InsideOut Jun 2 '16 at 13:18
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Complete a basis for $\mathbb R^2$ with $(-2,1)$.

Then $(x,y) \mapsto -2x+y$ is a linear map $\mathbb R^2 \to \mathbb R$ with kernel $\langle(1,2)\rangle$.

This map is the orthogonal projection onto the orthogonal complement of $\langle(1,2)\rangle$.

The induced map $\mathbb{R}^2/\langle(1,2)\rangle \to \mathbb R$ is $(x,y) \bmod \langle(1,2)\rangle \mapsto -2x+y$. (Make sure you understand that this map is well defined.)

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  • $\begingroup$ But the domain must be $\mathbb{R}^2/\langle(1,2)\rangle$ not $\mathbb{R}^2$ How would the explicit form of the linear transformation T? $\endgroup$ – Daniela Rondón Jun 2 '16 at 12:52
  • $\begingroup$ @DanielaRondón, see my edited answer. $\endgroup$ – lhf Jun 2 '16 at 13:17
  • $\begingroup$ Sorry what does it mean $(x,y) \bmod \langle(1,2)\rangle$? $\endgroup$ – Daniela Rondón Jun 2 '16 at 13:36

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