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Check if it's true or false:

$ \text{a) }f: R^2 \rightarrow R , f(x,y) = x\cdot y \text{ is linear}\\ \text{b) }f: V \rightarrow W \text{ and } dim(V)=dim(W) \text { so f is invertible}\\ \text{c) }\text{Exists a linear transfromation that maps }R^2 \text{ to } R^3 \text{ that is onto} $

For a) $$ f((x,y)+(a,b)) = f(x+a,y+b) = (x+a)(y+b)\\ f(x,y) + f(a,b) = xy + ab \\ f((x,y)+(a,b)) \neq f(x,y) + f(a,b) $$ FALSE

For b) $$ f:R^2 \rightarrow R^2, f(x,y) = (x-y,x-y)\\ ker(f) = span\left\{ (1,1) \right\}\\ dim(ker(f)) \neq \left\{ 0 \right\} \\ $$ Since it's not one to one, I can affirm that it's not isomorphic, therefore, not invertible.

For c) $$ f:R^2 \rightarrow R^3, f(x,y) = (x,y,c) \\ \text {or }c = 0 \text { or it is a combination between x and y} $$ in both cases for c, you can't get the third coordinate to vary throughout the real numbers independently from x and y, therefore, the image of that transformation will have some values that are not linked to it's domain, concluding that it's not onto.

--UPDATE (FOR C)-- $$ dim(ker(f)) + dim(im(f)) = 2 $$ If my transformation is injective, the dimension of it's kernel is going to be 0, therefore, I'll have $dim(im(f)) = 2$, concluding that a basis for R^3 can't be a span of two L.I (vectors basis for im(f)), I can guarantee that it's false.

It's correct?! All three are falses affirmatives??

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  • $\begingroup$ In b), I think you mean $ker(f)=span\{(1,1)\}$ instead of $\{(1,1)\}$. For c), you have to show in full generality that there exists no onto linear map $\mathbb{R}^{2}\rightarrow\mathbb{R}^{3}$. $\endgroup$
    – studiosus
    Commented Sep 15, 2016 at 13:06
  • $\begingroup$ I'll Edit that. Thanks!!! $\endgroup$
    – Bruno Reis
    Commented Sep 15, 2016 at 13:09

1 Answer 1

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You are correct that all three statements are false.

Your proof that a) is false is a little sloppy, because you assume it's obvious that $(x+a)(y+b) \not\equiv xy + ab$. It's sufficient, and customary, to choose specific vectors as counterexamples. So instead, if you could show $f(1,0) + f(0,1) \neq f(1,1)$, that would be cleaner.

Your counterexample for b) is good.

For c), you have made extra assumptions about $f$. What you want to show is that there does not exist an onto map $f$ from $\mathbb{R}^2$ to $\mathbb{R^3}$. In other words, every map $f$ from $\mathbb{R}^2$ to $\mathbb{R^3}$ is not onto. I"m not sure of your context, but if you're allowed to use dimension counts this is easy to show.

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  • $\begingroup$ Thanks for the answer my friend. For c, am I wrong? An answer using dimensions would be using rank-nullity, and with that I would see that a basis for image could only be <= 2, therefore, it would expand in maximum R^2, considering that the dim of kernel being 0... $\endgroup$
    – Bruno Reis
    Commented Sep 15, 2016 at 13:08
  • $\begingroup$ could you please check it now? $\endgroup$
    – Bruno Reis
    Commented Sep 15, 2016 at 13:13
  • $\begingroup$ @BrunoReis: looks good. The general statement is that if $f\colon V \to W$ is surjective, then $\dim W \leq \dim V$. $\endgroup$ Commented Sep 15, 2016 at 13:18
  • $\begingroup$ Thanks!! rank-nullity is very usefull! $\endgroup$
    – Bruno Reis
    Commented Sep 15, 2016 at 13:20

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