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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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    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

3 Answers 3

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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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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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    @ArturoMagidin Right on the money.2011-09-18
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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.