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Find the matrix $A$ of a linear transformation $T:\mathbb{R}^2\to\mathbb{R}^2$ that satisfies $$T\left[2\choose3\right] = {1\choose1}, \ T^2\left[{2\choose3} \right]= {1\choose2}.$$

I am trying to review some linear algebra and was confused about this question.

The answer given is $$A = \begin{pmatrix} 2 & -1 \\ 5 & -3 \end{pmatrix}$$ and I am not sure how it was obtained.

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  • $\begingroup$ With respect to which basis?? $\endgroup$ – Timbuc Feb 8 '15 at 8:54
  • $\begingroup$ @Timbuc: I think anyone can guess that to give the matrix of a linear transformation $\Bbb R^2\to\Bbb R^2$ one uses the standard basis of $\Bbb R^2$ (unless another one is specified). $\endgroup$ – Marc van Leeuwen Feb 8 '15 at 9:28
  • $\begingroup$ @MarcvanLeeuwen I prefer to ask and not to guess, in particular because right now I'm tutoring a course where many times it is required to give the matrix wrt non-canonical basis. $\endgroup$ – Timbuc Feb 8 '15 at 11:45
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Since it must be

$$T^2\binom23=T\left(T\binom23\right)=T\binom11=\binom12$$

we get, wrt the standard basis (BTW, check $\;\left\{\;\binom23\;,\;\;\binom11\;\right\}\;$ (check this is actually a basis!), first that

$$\begin{align}\binom10&=(-1)\binom23+3\binom11\\{}\\ \binom01&=1\cdot\binom23+(-2)\binom11\end{align}$$

we get that

$$\begin{align}&T\binom10:=T\left(-1\binom23+3\binom11\right)=-1\binom11+3\binom12=\binom25=2\binom10+5\binom01\\{}\\ &T\binom01=T\left(1\binom23-2\binom11\right)=\binom11-2\binom12=\binom{-1}{-3}=(-1)\binom10+(-3)\binom01\end{align}$$

and I thus get the matrix

$$\begin{pmatrix}2&\!\!-1\\5&\!\!-3\end{pmatrix}$$

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  • $\begingroup$ @Marc Thank you for the editing. $\endgroup$ – Timbuc Feb 8 '15 at 11:41
  • $\begingroup$ Timbuc, quick question. What if $\;\left\{\;\binom23\;,\;\;\binom11\;\right\}\;$ is not a basis? $\endgroup$ – Robben Feb 8 '15 at 22:55
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    $\begingroup$ @Robben Then $\;T\;$ is not uniquely defined. $\endgroup$ – Timbuc Feb 9 '15 at 1:45
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The linear transformation $T$ is defined by

$$T((2,3)^T)=(1,1)^T\quad\text{and}\quad T((1,1)^T)=(1,2)^T$$

since $B=((2,3)^T,(1,1)^T)$ is a basis for $\Bbb R^2$. Let $P$ the matrix change from the standard basis to $B$ then

$$P=\begin{pmatrix}2&1\\3&1\end{pmatrix}$$ so the matrix of $T$ relative to the standard basis is given by

$$[T]=\begin{pmatrix}1&1\\1&2\end{pmatrix}P^{-1}=\begin{pmatrix}2&\!\!-1\\5&\!\!-3\end{pmatrix}$$

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T^2[(2,3)] = (1,2) implies that T(T(2,3)) = (1,2) i.e. T(1,1) = (1,2).

The general linear transformation is not defined here. However, two linearly independent vectors are given. So, any general say, (x,y) vector can be given by the linear combination of these two vectors.

(x,y) = a*(2,3) + b*(1,1) -- (I)

On comparing we will get two equations in two unknowns x = 2a + b and y = 3a + b. From these two equations we can find the values of a and b in terms of x,y. Now substitute these in (I). Thus, we get the general co-ordinate and hence the general linear transformation as,

T(x,y) = aT(2,3) +bT(1,1) = (-x-y)(1,1) + (3x-2y)(1,2).

Now we can find the matrix of linear transformation with respect to the standard basis of R^2.

Hope this helps!

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