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I can't get my head around this introductory theorem from Lang's Linear Algebra.

Let $V$ and $W$ be vector spaces and $\{ v_1,\cdots v_n\}$ be a basis of $V$ and $w_1,\cdots , w_n$ be arbitrary elements of $W$. Then there exists a unique mapping $T:V\rightarrow W$ such that $T(v_i) = w_i $ for all $i$ from $1$ to $n$.

Let $V$ and $W$ be $\mathbb{R}^3$ and $w_1, w_2 ,w_3$ be three collinear points, then how can we have a mapping which maps the basis to these three points? Or am I missing something?

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    Of course you can; linear transformations have to respect linear dependencies *in the domain*, but the images can satisfy linear dependencies in the codomain that the original vectors did not. For example, what happens if you map everything to $0$; isn't that a linear transformation? The theorem tells you that (i) knowing what happens to a basis tells you exactly what happens to every vector in the domain; and (ii) you can decide what happens to the basis arbitrarily (i.e., *anything* you want can be done to the basis). Categorically, it says the vector space is free on the basis.2011-09-17
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    Reference: Page 95 , 2nd Ed. Page 56 in third edition, here is the [preview](http://books.google.com/books?id=0DUXym7QWfYC&lpg=PP1&pg=PA56#v=onepage&q&f=false)2011-09-17
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    @Arturo what do you mean by: "... vector space is free on the basis"2011-09-17
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    "free on the basis" there is a categorical notion. If you don't know categories, don't worry about it. (But essentially it means exactly what the theorem says: that you can decide "freely" what happens to the basis, and you can always find a way to extend that to a linear transformation from the *entire* space, and do so in a unique way).2011-09-17
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    I believe that this property is important, if a vector space have a set that verifies this, the set is a basis.2011-09-18
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    Let me mention one unrelated point. I think you are worried that the linear map will not be bijective or invertible if the basis vectors are all mapped to collinear points. It is true that it won't be invertible, but we are talking only about linear transformations that are not necessarily invertible.2011-09-18

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