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Find all linear transformations $T:\mathbb R^2\to \mathbb R^2$ such that $$T(1,2)=(2,1)\;\;\&\;\; T(2,5)=(1,3)$$

Are we supposed to find A? Since we are given x and the solution but I don't know how to find A such that A times (1,2) = (2,1), inverse perhaps?

Thank you!

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    $\begingroup$ Hint: Are the vectors $(1,2)$ and $(2,5)$ linearly independent? If so...then any $\vec v\in \mathbb R^2$ can be written as $\vec v=\lambda (1,2)+\mu (2,5)$ so $T(\vec v)$ would equal... $\endgroup$
    – lulu
    Jul 4, 2016 at 22:54
  • $\begingroup$ You'd better tell us what you call $A$ and not let us guess. $\endgroup$
    – user65203
    Jul 5, 2016 at 9:52

4 Answers 4

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Every single $T(a,b)$ can be expressed in the form: $T(a,b)=aT(1,0)+bT(0,1)$ $$T(0,1)=T(2,5)-2T(1,2)=(1,3)-2(2,1)=(1,3)-(4,2)=(-3,1)=T(\mathbf{e_2})$$ $$T(1,0)=T(1,2)-2T(0,1)=(2,1)-2(-3,1)=(2,1)-(-6,2)=(8,-1)=T(\mathbf{e_1})$$ So the standard matrix of $T$ is $A=\begin{bmatrix}8&-3\\-1&1\end{bmatrix}$, where $T(\mathbf v)=A\mathbf v$

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Hint: $\{(1,2),(2,5)\}$ is a basis of $\mathbb{R}^2$. As a linear map is uniquely determined by its action on a basis, there is precisely one linear map satisfying the prescribed conditions.

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Too long for a comment... From Rachel's answer one may easily derive an explicit formula for $T$, which is indeed unique as Alex stated in his answer.$$\\$$ Let $$v= \begin{bmatrix} v_1 \\ v_2 \end{bmatrix}$$ and just multiply Rachel's matrix $A$ (which is matrix representation of $T$) with $v$ so you'll get $$A \times v= \begin{bmatrix}8v_1-3v_2 \\-v_1+v_2\end{bmatrix}\\$$Now you know $$T(v)=T(v_1,v_2)=(8v_1-3v_2, -v_1+v_2)$$ $\forall v \in \mathbb R ^2$.

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In matrix form, you know that

$$A\left(\begin{matrix}1\\2\end{matrix}\right)=\left(\begin{matrix}2\\1\end{matrix}\right)$$ and $$A\left(\begin{matrix}2\\5\end{matrix}\right)=\left(\begin{matrix}1\\3\end{matrix}\right)$$

which you can regroup as $$A\left[\begin{matrix}1&2\\2&5\end{matrix}\right]=\left[\begin{matrix}2&1\\1&3\end{matrix}\right].$$

Can you take it from here ?

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