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How to quickly and clearly argue/show that there is or is not a linear surjective map $\phi$

$\phi: \mathbb{R} \to \mathbb{R}²$

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    Any thoughts yourself? How would the matrix of such a map look like? If $f(a)=(0,1)$ for some $a\in\mathbb R$, then what might map to $(1,0)$?2012-02-01
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    @Listing: First of all, there *is* a continuous surjective map from $\mathbb{R}$ to $\mathbb{R}^2$: you seem to have misread the link you quoted. (There is no continuous *bijection*.) Second of all, the OP has asked just about the most elementary possible linear algebra question: in view of this, it's difficult for me to believe that your remark will be helpful or even meaningful to him or her.2012-02-01
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    It looks like that there **is not** such a map. But I don't know a waterproof way of showing this. It's just that I can't imagine how there should be such a map, but this is not enough to say that there is not such a map at all.. I think it's not hard to see that there can't be such a map, but how to argue?2012-02-01
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    @meinzlein: I'm not sure whether you are asking about the nonexistence of a continuous surjection (in which case I retract some of my previous comment) or asking again about the linear case (in which case, I would direct you to my answer below and ask you to let me know if you have further questions). Regarding the former: it is well-known that there are continuous surjective maps $f: [0,1] \rightarrow [0,1]^2$. Using the fact that we can fit countably many disjoint closed intervals into the real line and cover the plane by countably many closed squares, the result follows.2012-02-01
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    @PeteL.Clark Sorry for the stupid mistake, I did not know there was a continuous surjection from R to R^2. However I added the remark to link the questions because someone who stumbles on this question using a search engine might be interested in the second similar question. Although I agree that it will not be helpful for the OP.2012-02-02

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Assuming you mean an $\mathbb{R}$-linear map, one can argue as follows: let $v = \varphi(1)$. Then for all $\alpha \in \mathbb{R}$, $\varphi(\alpha) = \varphi(\alpha \cdot 1) = \alpha \varphi(1) = \alpha v$. Thus the image of $\varphi$ consists of all scalar multiples of the vector $v$. This is the origin if $v = 0$ and a line through the origin otherwise. It is clearly not the entire plane: e.g., if $v = (x,y)$ is not zero, then $(-y,x)$ is not of the form $\alpha (x,y)$ for any $\alpha \in \mathbb{R}$.

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For a linear transformation in this case we have: 1=dim R =dim ImT + dim KerT. Then dim ImT is 0 or 1.