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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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2 Answers 2

up vote 0 down vote accepted

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 $W$? –  jpp 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 $\mathbb{R}^2 \to \mathbb{R}^2$. –  jpp Mar 27 '13 at 16:59
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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.

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I was actually going to include this as another way I thought of solving the problem. I'd still like something a little more rigorous though. –  jpp Mar 27 '13 at 3:16
    
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
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@user1850672: As Branimir said, there’s nothing unrigorous about noticing an example and verifying it. I suspect that you really mean that you’d like something a little more systematic, something that doesn’t rely on spotting an example; DonAntonio’s answer gives you that. –  Brian M. Scott Mar 27 '13 at 3:46
    
@BrianM.Scott Yes I apologize, systematic was the word I was looking for. Thank you for your help! –  jpp Mar 27 '13 at 4:43
    
@user1850672: You’re welcome! –  Brian M. Scott Mar 27 '13 at 4:44

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