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Let $V,W$ two vector spaces over the field $F$.

If $x_1, x_2\in V$ and $y_1, y_2\in W$, then there exists a linear transformation $T: V\to W$ so that $T(x_1)=y_1$ and $T(x_2)=y_2$.

I think that statement is false, but, I do not know how to prove it.

Thanks for your help!

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    $\begingroup$ Try to define a linear transformation, $T: \mathbb{R} \to \mathbb{R}$, such that $T(1) = 2$, but $T(2) = 5$. $\endgroup$ – Joe Johnson 126 Sep 30 '17 at 13:17
  • $\begingroup$ When $x_1=x_2$, and $y_1\neq y_2$ it is certainly impossible to find such a $T$. $\endgroup$ – Mike Earnest Sep 30 '17 at 17:44
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Without further restrictions on $x_1$, $x_2$, $y_1$, $y_2$ the statement in general is not true. For example, if $x_2=\lambda x_1$ but $y_2\ne\lambda y_1$ then obviously you can not find $T$. However, if $x_1$ is linearly independent from $x_2$ and similarly $y_1$ is linearly independent from $y_2$, then you can always find a $T$, and if the space has dimension larger than 2 there are infinitely many solutions.

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  • $\begingroup$ If $x_1, x_2$ are linearly independent, then linear independence of $y_1, y_2$ is irrelevant. The basic theorem is: if $\{x_1, \dots, x_n\}$ is a basis of $V$, and $y_1, \dots, y_n \in W$ are arbitrary, then there exists a unique linear transformation $T: V \to W$ such that for all $i$, $T x_i = y_i$. $\endgroup$ – fredgoodman Sep 30 '17 at 16:28

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