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Just as a prelim. to this question, we have a from our book that states:

Theorem 2.6

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 $T\colon V \to W$ such that $T(v_i) = w_i$, for $i = 1, 2, \dots, n$.

So im stumped on this problem here.

Assume that $V$ and $W$ are both finite-dimensional vector spaces over $F $and $T$ is a function from $V$ to $W$.

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

Give a counterexample on why this statement is false.

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    From which book is this question taken? – 2017-02-28
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    Linear algebra 3rd edition, stephen H friedberg, arnold j insel, and lawrence e spence – 2017-02-28
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    @PraneetSrivastava I came across the same question in Linear Algebra by A.R. Rao and P. Bhimasankaram! – 2018-12-25

2 Answers 2

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For example, you may take $x_1=0$ and $y_1\neq0$. Then there is no linear transformation T such that $T(x_1)=y_1$, because $T(x_1)=T(0)=T(x_2-x_2)=T(x_2)-T(x_2)=0\neq y_1.$

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Also, it suffices to show that if $x$ and $x_1$ are not linearly independent then $T(x)$ is a scalar multiple of $T(x_1)$ and thus it's really easy to construct a counterexample.