# Determining if linear transformation is well defined

True or False: There is a linear transformation $T$: $\mathbb{R}^{2} \to \mathbb{R}^{2}$ with

$$T \left(\begin{array}{c} 1\\ 2\\ \end{array} \right) = \left(\begin{array}{c} 1\\ 0\\ \end{array} \right)\quad\text{ and }\quad T \left(\begin{array}{c} 2\\ 1\\ \end{array} \right) = \left(\begin{array}{c} 2\\ 0\\ \end{array} \right)\;.$$

I know one way to verify this statement is to form a system of equations with a general matrix $\left(\begin{array}{c c} a & b\\ c & d\\ \end{array} \right)$ and those mappings and solve and see that it is well-defined?

How can I verify this using theory about linear transformations?

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The theorem that solves this and any other similar problem is:

Theorem: Let $\,V\,$ be a vector space of dimension $\,n\,$ and let $\,\{v_1,\ldots,v_n\}\,$ be some basis of it. Let $\,W\,$ be any other vector space over the same field as $\,V\,$ and let $\,\{w_1,\ldots,w_n\}\,$ be any $\,n\,$ vectors in $\,W\,$ . Then there exists a unique linear transformation $\,T:V\to W\,$ s.t $$\,Tv_i=w_i\;,\;\;\forall\,i=1,2,...,n\,$$

This theorem solves your problem at once after observing that

$$\left\{\;\binom{1}{2}\;,\;\binom{2}{1}\;\right\}$$

is a basis of $\,\Bbb R^2_{\Bbb R}\,$

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Then do these $w_i$'s form a basis for the vector space $W$? –  jp24 Mar 27 '13 at 4:01
@user1850672 They don't have to; indeed, the $w_k$ need not even be distinct. In any event, the failure of $\{w_k\}$ to be a basis for $W$ is precisely the failure of $T$ to be surjective. –  Branimir Ćaćić Mar 27 '13 at 4:04
@BranimirĆaćić Intuitively why is $T \left(\begin{array}{c} 1\\ 2\\ \end{array} \right)$ (which is clearly not a basis for $\mathbb{R}^2$) $\mapsto \left(\begin{array}{c} 1\\ 0\\ \end{array} \right)$ not enough to define a linear transformation from $\mathbb{R}^2 \to \mathbb{R}^2$. –  jp24 Mar 27 '13 at 16:59
Because you can't extend the definition to all the elements of $\,\Bbb R^2\,$ , for example: what would be $\,T\binom{0}{1}\,$ ? –  DonAntonio Mar 27 '13 at 17:09
I would do this one by inspection: the transformation $T\binom{x}y=\binom{x}0$ clearly has the desired properties.
As long as you check to your satisfaction that $T : \mathbb{R}^2 \to \mathbb{R}^2$ is a linear transformation, there's nothing the least bit unrigorous about proving the existence of the desired linear transformation by pointing to this explicit example. –  Branimir Ćaćić Mar 27 '13 at 3:37