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Let $D$ be a square with vertices $(0, 0), (1, 1), (2, 0), (1, −1)$ and $D^*$ be a parallelogram with vertices $(0, 0), (1, 2), (2, 1), (1, −1)$. Find a linear map T taking $D^*$ onto $D$.

The answer is $$T(x,y) = (x, (-x/3)+(2x/3)),$$ but I'm not sure how to get that?

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Hint: Based on what they tell us, it suffices to take a $T$ satisfying $T(1,1)=(1,2)$ and $T(1,-1)=(1,-1)$. This is enough for us to deduce what $T$ is.

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    why is that enough, and not all 4? – 2017-02-02
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    @user3427042 If there is such a linear map at all, the fourth point is redundant: $(1,-1)=(2,0)-(1,1)$, so it provides no additional information. You should check, however, that T(1,-1)=T(2,0)-T(1,1)=(1,-1) – 2017-02-02