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.


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

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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