Why doesn't this linear transformation exist? In Friedberg's Linear Algebra, the following theorem is presented:
"Let V and W be vector spaces over F, and suppose that V is finite-dimensional with a basis $\left\{v_1,...,v_n\right\}$. For any vectors $w_1,...,w_n$ in W there exists exactly one linear transformation $T:V\rightarrow W$ such that $T(v_i)=(w_i)$ for $i=1,...,n$."
I think I understand what this is saying, but then in the exercises there is a true/false question that is supposedly false that is as follows:
"Given $x_1,x_2\in V$ and $y_1,y_2\in W$, there exists a linear transformation $T:V\rightarrow W$ such that $T(x_1)=y_1$ and $T(x_2)=y_2$."
My question is why is this false? Doesn't it follow from the theorem?

Let $V=W=\mathbb{R}^2$. Let $x_1 = (1,0), x_2 = (2,0), y_1=(0,1), y_2=(1,3)$. Suppose there exits a linear transformation $T$ as required. Then
$$T(2,0)=(1,3).$$ Also $$T(2,0)=2T(1,0)=2(0,1)=(0,2).$$
This gives a contradiction. The reason this happens, as mentioned in the comments, is that $x_1$ and $x_2$ are linearly dependent.
