Given $\alpha_{i}$ as well as $\beta_{i}$, is there a linear transformation such that $T(\alpha_{i}) = \beta_{i}$?

Question

If $\alpha_{1} = (1,-1)$, $\alpha_{2} = (2,-1)$ and $\alpha_{3} = (-3,2)$ as well as $\beta_{1} = (1,0)$, $\beta_{2} = (0,1)$ and $\beta_{3} = (1,1)$, is there a linear transformation $T$ from $\textbf{R}^{2}$ into $\textbf{R}^{2}$ such that $T(\alpha_{i}) = \beta_{i}$?

MY ATTEMPT

To start with, let us determine the expression of $T$ based on the values of $\alpha_{1}$ and $\alpha_{2}$. Given that $(x,y)\in\textbf{R}^{2}$, it can be rewritten as \begin{align*} (x,y) = -(x+2y)(1,-1) + (x+y)(2,-1) \Longrightarrow T(x,y) = (-x-2y,x+y) \end{align*} Based on such results, one has $T(\alpha_{1}) = T(1,-1) = (1,0) = \beta_{1}$, $T(\alpha_{2}) = T(2,-1) = (0,1) = \beta_{2}$. Finally, $T(\alpha_{3}) = T(-3,2) = (-1,-1)$, from whence we conclude there is no possible way to construct such linear transformation.

My question is: is there another way to prove there is no such linear transformation based on theoretic results only? Any help is appreciated.

Answer 1

The first thing you need to ask yourself is: are $\alpha_1$, $\alpha_2$ and $\alpha_3$ linearly independent?

If they are, then you can map them into three arbitrary vectors $\beta_1$, $\beta_2$ and $\beta_3$ and you can extend that to a linear map by setting:

$$T(u\alpha_1+v\alpha_2+w\alpha_3)=u\beta_1+v\beta_2+w\beta_3$$

$(u,v,w\in\mathbb R)$.

If, however, they are linearly dependent, which means, for example, that $\alpha_3$ is a linear combination of $\alpha_1, \alpha_2$ (say, $\alpha_3=u\alpha_1+v\alpha_2$), then you cannot freely choose what to map $\alpha_3$ into - even if you can freely choose what to map $\alpha_1$ and $\alpha_2$ into. Namely, $T(\alpha_3)$ is bound to be:

$$T(\alpha_3)=T(u\alpha_1+v\alpha_2)=uT(\alpha_1)+vT(\alpha_2)$$

and is uniquely determined by what $\alpha_1$ and $\alpha_2$ map to.

In your case, $\alpha_1$ and $\alpha_2$ are linearly independent, but $\alpha_3=-\alpha_1-\alpha_2$. Therefore, for any linear map $T$ that maps $\alpha_1$ to $\beta_1$ and $\alpha_2$ to $\beta_2$ we would be bound to have:

$$T(\alpha_3)=T(-\alpha_1-\alpha_2)=-\beta_1-\beta_2=(-1, -1)\ne(1,1)=\beta_3$$

so $T(\alpha_3)$ cannot be $\beta_3$. Putting it differently, there is no such linear map $T$ that maps $\alpha_1$, $\alpha_2$, $\alpha_3$ into $\beta_1$, $\beta_2$, $\beta_3$, respectively.

Answer 2

The shorter way is: $\alpha_1+\alpha_2+\alpha_3 = 0$, but $\beta_2 + \beta_2 + \beta_3 \ne 0$. So there is no linear transformation mapping $T(\alpha_i) = \beta_i$for $i=1,2,3$.

Answer 3

we have $T(\alpha_{3})=\beta_{3}=\beta_{1}+\beta_{2}=T(\alpha_{1})+T(\alpha_{2})=T(\alpha_{1}+\alpha_{2})=T(-\alpha_{3})=-T(\alpha_{3})$ therefore $T(\alpha_{3})=(0,0) , (=\beta_{3})$ Which is impossible. Therefore such a linear map doesn't exist