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How can I prove that the transformation $$ T: R^3 \to R^2 : \ T(a_1, a_2, a_3) = (a_1-a_2, 2a_3)$$ is one-to-one and/or onto?

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What is the image of $(x,0,y/2)$? And of $(x,x,z)$? –  Sigur Oct 16 '12 at 23:18
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@Sigur: perhaps $(x+z,z,y/2)$ might have been another educational example –  Henry Oct 16 '12 at 23:23
    
To see it is not one to one, just check the image of the two points $(1,1,1)\,,(-1,-1,1)$. These two points shares the same image $ (0,1) \,,$ which implies that the map is not one to one. –  Mhenni Benghorbal Oct 17 '12 at 1:02
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For one-to-one you need to show that if two points in $\mathbb R^3$ are mapped to the same point, they are the same. So let $x=(x_1,x_2,x_3)$ and similarly for $y$. Then assume $T(x)=T(y)$, insert the definition of $T$ and see if you can prove that $x_1=y_1$ and so on.

For onto, you need to show that all points in $\mathbb R^2$ can be reached. So let $x=(x_1,x_2)$ and see if you can find a $y$ such that $T(y)=x$. You should be able to find many of them. If you do this part first, it answers the other as well.

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Is there a general way to prove it, i.e. for $R^n \to R^m$? –  Imray Oct 16 '12 at 23:35
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@Imray: If $n \gt m$, it cannot be one-to-one. If $n \lt m$, it cannot be onto. You should soon learn a theorem about the sum of the dimensions of the kernel and image adding to the dimension of the domain. –  Ross Millikan Oct 16 '12 at 23:39
    
I can see from the dimension theorem how you would prove when n < m, but how do you prove it cannot be 1-to-1 when n > m? In the equation $rank(T)+Nullity(T) = dim(V)$, Nullity can be equal to zero. –  Imray Oct 17 '12 at 3:51
    
@Imray: Because the rank(T) can't be greater than dim(range), not dim(domain) –  Ross Millikan Oct 17 '12 at 3:55
    
I still dont understand. By the dimension theorem, for $T: V \to W, \ R^n \to R^m, \ \ rank(T)+nullity(T)=dim(v)$. So when $n < m $, I can clearly see for the formula to add up (with $rank(T) = m$) that $nullity(T) > 0$ which means the transformation cannot be one-to-one. But what about when $n>m$? –  Imray Oct 17 '12 at 13:17
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