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Prove or disprove statement Given $x_1,x_2\in V$ and $y_1,y_2 \in W$, $\exists$ a linear transformation $T:V\to W $ such that $T(x_1)=y_1$ and $T(x_2)=y_2$


Relate Thm

Let $V$ and $W$ be vector spaces over $F$, and suppose that $V$ is finite-dimensional with basis $\{v_1,..,v_n\}$. For any vectors $w_1,..,w_n \in W$ there exists exactly one linear transformation $T:V\to W$ such that $T(v_i)=w_i$ for $i=1,\dots,n$


So, I know it should be false. Guessing the trick is to let $V$ and $W$ not be Vector Spaces Then T cannot be linear


I found from a solutions from TA that

if $x_2=2x_1$ then $T(x_2)$ must be $2T(x_1)=2y_1$

I guess it is shown here that there a $y_1$ and $y_2$ are related by a multiple. I can't see how that statement along will disprove the statement.

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  • $\begingroup$ Why do you know that "it should be false"? If $V$, $W$ are not vector spaces, the statement does not make sense. $\endgroup$ Commented Feb 18, 2016 at 14:42
  • $\begingroup$ It has false on the solution on back of the book. $\endgroup$ Commented Feb 18, 2016 at 14:45
  • $\begingroup$ Does it say in the book (or on the back), what we assume on $V$ and $W$ ? $\endgroup$ Commented Feb 18, 2016 at 14:46
  • $\begingroup$ on the back there is just "False". $\endgroup$ Commented Feb 18, 2016 at 14:47

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First of all you know V and W to be vector spaces, or you can't even define a linear transformation between them. I'll assume that they are. Your statement is not always true, in fact you can choose $y_1$ and $y_2$ as you like if and only if $x_1$ and $x_2$ are linearly independent. For this to happen necessarily $dim V \ge2$ and your example shows in fact that it is not possible in $\mathbb{R}$, that has dimension 1 as a vector space. However, chosen $V$ with greater dimension and $x_1,x_2$ linearly independent (meaning, for two vectors, not one multiple of the other) you can choose whatever $y_1,y_2$ you like.

EDIT if $x_1,x_2$ are one multiple of the other you can freely choose $y_1$ or $y_2$, and the other will be determined by the linearity of $f$.

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