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Is there a linear transformation $T : R ^2 → R ^3$ such that $T(1, 1) = (1, 0, 2)$ and $T(2, 3) = (1, −1, 4)$. Justify your answer.

I'm not sure what exactly this question is asking for. How would you go about tackling this question. I know I'm supposed to show some form of working on this website but I really have no idea where to start

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    $\begingroup$ Try to start with the assumption that there is such a transformation. Then try to explicitly construct it by finding a matrix $A$ such that $T(x) = Ax$ for all $x\in \Bbb R^2$. If you're able to, then one clearly exists. If you run into some contradiction, then one clearly doesn't exist. $\endgroup$ – user137731 Oct 4 '15 at 4:25
  • $\begingroup$ As always, reviewing the terms in the problem (what a linear transformation is) can help. $\endgroup$ – Nate 8 Oct 4 '15 at 4:30
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In general if V and W are vector spaces and $\{e_i\}_{i\in I}$ is a basis for $V$. And $f(e_i)=w_i\in W$ for all $i \in I$.

Then f can be extended to a linear mapping $T: V \rightarrow W$ such that $T(e_i)=f(e_i)$. (You can write down the proof in the same way as the above answers)

In your case, $V= \mathbb R ^2 $ and $W=\mathbb R^3$ $e_1=(1,1), e_2=(2,3)$ is a basis for $\mathbb R^2$ and $f(e_1)=(1,0,2)$ and $f(e_2)=(1,-1,4)$.

Therefore, you can extend it to a linear mapping by the above discussion.

Note: 1) $(1,0,2)$ and $(1,-1,4)$ doesn't have any significance role to play.If $(a,b,c), (d,e,f)$ still you could extend it to a linear map.

2) The above problem has a general philosophy - any map (or morphism) from a algebraic(mathematical) structure to another algebraic(mathematical) structure can be defined by only defining it on the generators, provided you can talk of generators. For example, to define a group homomorphism from a cyclic to group to another group, it is enough to define it for just for the generators.

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Note that $\{(1,1),(2,3)\}$ is a basis of $\mathbb R^2$ and if you know the image of basis elements then you know the whole transformation.

In this case observe $\forall x,y \in \mathbb R^2$ we have $(x,y)= (3x-2y)(1,1)+(y-x)(2,3)$

So define , $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$

To answer the comment by Babai:

The construction of $T$ shows that if $u\in \mathbb R^2,u=\alpha a+\beta b$ where $a=(1,1)$ and $b=(2,3)$ then $Tu=\alpha Ta+\beta Tb$

Thus $T(u_1+\lambda u_2)=T(\alpha_1 a+\beta_2 b+\lambda (\alpha_2 a+\beta_2 b))=T((\alpha_1+\lambda \alpha_2) a+(\beta_1+\lambda \beta_2) b)=(\alpha_1+\lambda \alpha_2) Ta+(\beta_1+\lambda \beta_2) Tb=\alpha_1 Ta+\beta_2 Tb+\lambda (\alpha_2 Ta+\beta_2 Tb)=Tu_1+\lambda Tu_2$

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  • $\begingroup$ The question is why is T linear? In your answer you assumed T is linear. $\endgroup$ – Babai Oct 4 '15 at 4:41
  • $\begingroup$ @Babai I have added it in my answer. $\endgroup$ – usermath Oct 4 '15 at 5:22
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    $\begingroup$ $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$. Why is this true? This is only true if $T$ is linear, and that is what you nee to prove. $\endgroup$ – Babai Oct 4 '15 at 5:38
  • $\begingroup$ Instead of writing " So $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$, if you write "define $ T$ as $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$, then your answer will make perfect sense. $\endgroup$ – Babai Oct 4 '15 at 5:45
  • $\begingroup$ @Babai ok I got it and corrected. Thnx $\endgroup$ – usermath Oct 4 '15 at 6:27
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Define $T: \mathbb{R}^2 \to \mathbb{R}^3$ by for $(x,y) \in \mathbb{R}^2$, write $(x,y) = a(1,1) + b(2,3)$, and let $T(x,y) = a(1,0,2) + b(1,-1,4).$

In the above, $T$ is obviously linear. The problem is why such a $T$ is well-defined, more specifically, why there exists such scalars $a,b$ and the scalars are unique.

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  • $\begingroup$ Why is T obviously linear? $\endgroup$ – Babai Oct 4 '15 at 4:42
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    $\begingroup$ @Babai Have you tried to show it before you ask this question? It is a one line proof. $\endgroup$ – Empiricist Oct 4 '15 at 5:17
  • $\begingroup$ It is not the question of whether I have tired it or not. Your answer is not justifiable. You write "$T(x,y) = a(1,0,2) + b(1,-1,4).$, In the above, T is obviously linear." Now, $T(x,y) = a(1,0,2) + b(1,-1,4).$ , this is not true unless $T$ is linear. And you assumed it to say $T$ is linear, which is not correct. $\endgroup$ – Babai Oct 4 '15 at 5:35
  • $\begingroup$ I am sorry. You are defining $T$. Your answer is perfect. And it is obviously linear. $\endgroup$ – Babai Oct 4 '15 at 5:44
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If there exists a transform, there must exist a matrix representing it, let us call such a matrix $\bf T$. Now what we demand is that $${\bf T}\left[\begin{array}{cc}1&2\\1&3\end{array}\right] = \left[\begin{array}{rr}1&1\\0&-1\\2&4\end{array}\right]$$ Now we can solve this, if we divide by $\left[\begin{array}{cc}1&2\\1&3\end{array}\right]$ to the left of each side: $${\bf T} = \left[\begin{array}{rr}1&1\\0&-1\\2&4\end{array}\right] \left[\begin{array}{cc}1&2\\1&3\end{array}\right]^{-1} = ?$$


Now we can see on the matrices that this must have a solution since both matrices are of rank 2.


Then after we have solved this, what remains is to perform the matrix multiplication ${\bf T}\left[\begin{array}{cc}1&2\\1&3\end{array}\right]$ and confirm it really becomes as the first equation says it should.

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