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Let $v_{1} = (1, 0)$, $v_{2} = (1, -1)$ and $v_{3} = (0, 1)$. How many linear transformations $T :\mathbb {R^2}\rightarrow \mathbb {R^2} $ are there such that $T(v_{1} ) = v_{2}$, $T(v_{2} ) = v_{3}$, $T(v_{3} ) = v_{1}$. I am finding difficulty in tackling to this problem. I tried to identify corresponding linear transformation. But didn't come to any conclusion.It should be either 0, 1, 3 or $3!$

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Can you find even one such transformation? – Chris Eagle May 9 '12 at 12:46
That's what i am asking. i think answer should be zero. But how to show? – srijan May 9 '12 at 12:48
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Remember that once you have defined a linear map on a basis, you can work out what its value has to be on any other vector. – Matt Pressland May 9 '12 at 12:49
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If you have some ideas about the problem (like that the answer is zero), then put them in your question. – Chris Eagle May 9 '12 at 12:49
i have added sir..it should be either 0 , 1 , 3 or 3! – srijan May 9 '12 at 12:51
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$v_2 = v_1-v_3$, so you'd need: $$ v_3 = T(v_2) = T(v_1-v_3)=T(v_1)-T(v_3) = v_2-v_1$$ which is not true.

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I think a hint would probably have been more helpful than a complete answer. (I'm just bothering to say this because I upvoted your answer, and then realised that I don't like the fact that complete answers to easy questions get disproportionately many upvotes, so upvoting it perhaps didn't make sense.) – Tara B May 9 '12 at 13:12
@Tara B: Unless the OP merely copies the above answer, there is work for the OP to do in order to see that the above does answer the question. – André Nicolas May 9 '12 at 13:31
@TaraB Actually, I agree with you, I probably should have just given a hint. Sometimes, my eagerness to answer overrides the filter that asks, "What kind of answer is best for this user?" I'll leave it as is, since I'll guess the OP has already read it, but will keep trying to improve my instincts for future answers. – Thomas Andrews May 9 '12 at 13:39
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@AndréNicolas: That's true, but I think there's still more benefit to be gained from working out the rest from a hint than from figuring out why an answer works. – Tara B May 9 '12 at 13:46

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