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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. –  Arturo Magidin Sep 17 '11 at 23:54
Reference: Page 95 , 2nd Ed. Page 56 in third edition, here is the preview –  kuch nahi Sep 17 '11 at 23:55
@Arturo what do you mean by: "... vector space is free on the basis" –  kuch nahi Sep 17 '11 at 23:56
"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). –  Arturo Magidin Sep 17 '11 at 23:58
I believe that this property is important, if a vector space have a set that verifies this, the set is a basis. –  Frank Murphy Sep 18 '11 at 3:59
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3 Answers

The following exercise will strengthen your understanding of the concept that you've asked about above.

Let $T$ be a linear transformation between two finite dimensional vector space $V$ and $W$ of dimensions $n$ and $m$ respectively. Prove that there exist bases $B$ of $V$ and $C$ of $W$ such that matrix of the linear transformation with respect to these bases (which we denote by $M_{C}^B(T)$ ) has the form:

$ M = \begin{pmatrix} I_r &|& 0 \\ \hline 0 &|& 0 \end{pmatrix} $

where $r$ is the dimension of the image of $T$. $I_r$ by the way is the $r \times r$ identity matrix.

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The main point is that a linear transformation is determined by its action on a basis. That's the essence of linearity.

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That's half the main point; the other half is that the action on a basis can be arbitrary. –  Arturo Magidin Sep 18 '11 at 3:54
@Arturo, right, as usual. –  lhf Sep 18 '11 at 4:01
@ArturoMagidin Right on the money. –  fpqc Sep 18 '11 at 8:09
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A quick and easy example. Take $w_1=(0,0,0), w_2=(1, 0, 0), w_3=(2, 0, 0)$. Define

$$f(x, y, z)=(y+2z, 0, 0).$$

This $f$ is linear and maps the basis $(e_1=(1, 0, 0), e_2=(0,1,0), e_3=(0,0,1))$ into $(w_1, w_2, w_3)$.

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