5

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

$\phi: \mathbb{R} \to \mathbb{R}²$

meinzlein
  • 447
  • 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 $(1,0)$? – hmakholm left over Monica Feb 01 '12 at 17:45
  • 8
    @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. – Pete L. Clark Feb 01 '12 at 17:51
  • 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? – meinzlein Feb 01 '12 at 18:06
  • @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. – Pete L. Clark Feb 01 '12 at 18:19
  • @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. – Listing Feb 02 '12 at 06:56

2 Answers2

13

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 $\alpha \in \mathbb{R}$.

Pete L. Clark
  • 97,892
6

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

alpha.Debi
  • 1,034