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I have to determine whether the function $f:\mathbb{R}^2\to\mathbb{R}^2$ defined by $f(x,y)=(x+y,x)$ is one-to-one, onto, and if both then describe its inverse. I'm relatively new to mapping since I've only had Linear Algebra so I am not sure how to do this problem at all. I've asked others and they've included equations and stuff like that but didn't tell me where they got them from. So basically I need a step by step explanation of this problem. Thanks.

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If I wrote down the map in matrix form, it's given in the standard basis like $$ \left( \begin{array}{c c} 1 & 1 \\ 1 & 0 \end{array} \right)$$. Now use your linear algebra! Is it injective/surjective? Try using a determinant. –  A Blumenthal Jan 24 '13 at 21:19

3 Answers 3

One-to-one: Let $(x_1,y_1),(x_2,y_2)\in\mathbb{R}^2$ and assume that $f(x_1,y_1)=f(x_2,y_2)$. Then $$ (x_1+y_1,x_1)=(x_2+y_2,x_2) $$ or in other words $(x_1+y_1)=(x_2+y_2)$ and $x_1=x_2$. Is this enough to deduce that also $y_1=y_2$, and hence $(x_1,y_1)=(x_2,y_2)$?

Onto: Let $(x,y)\in\mathbb{R}^2$. The goal is to find $(a,b)\in\mathbb{R}^2$ such that $f(a,b)=(x,y)$. But this is just to find $(a,b)$ such that $$ (a+b,a)=(x,y). $$ Can you find $a$ and $b$ from here?

Inverse: Let $(x,y)\in\mathbb{R}^2$ and consider the point $f(x,y)=(x+y,x)$. What do we have to do to $(x+y,x)$ in order to get back to $(x,y)$? Clearly, the second coordinate $x$ should be mapped to the first coordinate. And if we take the first coordinate $x+y$, subtract the second coordinate $y$ and map the result into the second coordinate we obtain $(x,y)$. Mathematically, this is given by $$ f^{-1}(x,y)=(y,x-y),\quad (x,y)\in\mathbb{R}. $$

Check: Let us check our calculations by verifying that $f^{-1}(f(x,y))=(x,y)$: $$ f^{-1}(f(x,y))=f^{-1}(x+y,x)=(x,(x+y)-x)=(x,y),\quad (x,y)\in\mathbb{R}. $$

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Why do so much work? –  Bob Jan 25 '13 at 1:03
    
@Bob: Why not? $\,$ –  Stefan Hansen Jan 25 '13 at 6:52
    
Cause it can be done soo much easier, not to mention quicker.. –  Bob Jan 25 '13 at 11:22
    
@Bob: So you're saying that an answer necessarily has to be the shortest path between problem and solution? Hopefully, someone else than you can appreciate longer solutions, when in fact the shortest solutions are given in other answers. –  Stefan Hansen Jan 25 '13 at 11:42
    
@Bob. I don't think an answer that uses the tools of linear algebra is that much shorter or quicker if equal level of preciseness is maintained. Besides, in a question like this it seems more likely that the questioner is checking whether the person is able to work around with the fundamentals. –  Thomas E. Jan 25 '13 at 12:09

Since you've had linear algebra, you should be able to show that the given function is linear, so it has a matrix in some bases. The standard basis would work in this case. Then you have to show that the matrix you get is non-singular. This can be done either by computing the determinant or the rank. Then you conclude that your function is invertible.

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Hint: See the representation matrix of the linear transformation $f$.

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