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Find a linear map $ f: \mathbb{R} ^3 \to \mathbb{R} ^2$ such that there exists $ A \subset \mathbb{R} ^3$ which is linearly dependent and $ f(A)$ is linearly independent.

The only thing I thought is that $f$ is not injective. But I can't find such an $f$.

Thank's in advance.

2 Answers 2

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Take $A= \{(1,0,0), (0,1,0),(\frac{1}{2},\frac{1}{2},0)\}$ and define $F$ as following $F(1,0,0)= (1,0)$ $F(0,1,0)= (1,0)$ $F(0,0,1)= (0,1)$ Extend it lineary; That is $F(x,y,z)= xF(1,0,0)+ yF(0,1,0)+ zF(0,0,1)$ $F(x,y,z)= (x+y, z)$ Then $F(A)=\{(1,0)\}$ which is independent.

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    @sebastian, Yeah you are right.. My mistake.. editing it.. thanks ..2012-02-22
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This is a complete revision of my first (terrible) answer.

$\{\mathbf x\}$ is a perfectly good linearly independent set in $\mathbb R^2$ when $\mathbf x = (x_1,x_2) \neq0$. Consider a $\mathbf y \in \mathbb R^3$ which is orthogonal to $(x_1,x_2,0)$ then $A = \{(x_1,x_2,0)+a\mathbf y\,|\,a\in\mathbb R\}$ is linearly dependent and the orthogonal projection onto $\operatorname{span}\{\mathbf x\}$ has the desired property since all elements in $A$ get mapped to $\mathbf x$.

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    @passenger What do you mean by $f \equiv x$?2012-02-22