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How would I show $f\colon(1,0,3),(0,2,0),(1,2,3) \to (1,3),(0,2),(1,5)$ is unique.

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    Do you know yet that linear maps are uniquely determined by their action on a basis? This is ultimately what you should use, but you may have to prove it in this case.2012-12-05

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The first two vectors $(1,0,3)$ and $(0,2,0)$ are linearly independent, and span a two dimensional subspace $W$ of $R^3$. The third vector $(1,2,3)$ is an element of $W$ since it is the sum of the first two vectors.

You can say that there is a unique linear map $L$ taking $(1,0,3)$ to $(1,3)$ and also taking $(0,2,0)$ to $(0,2)$. Then from linearity you would have $L(1,2,3)=L(1,0,3)+L(0,2,0)=(1,3)+(0,2)=(1,5).$ That is, your third requirement that $(1,2,3)$ goes to $(1,5)$ is a linear consequence of the prescription of where the first two vectors go, if the result is to be a linear map.