Definition of a specific linear transformation being applied to create a change of variable matrix - Example - Lin Alg by Friedberg I have a question about a specific example of a linear transformation used to define a change of coordinate matrix. The example comes from Linear Algebra by Friedberg, Insel, et.al. The following is a screenshot of the example:


My question pertains to what linear transformation ,$T$, are they using to define the basis $\beta'$? The trouble began in trying to figure out how $T(-2,1) = -(2,1) = (2,-1)$ and how this applies to the rotation of the axis that the authors are trying to accomplish. Would someone be able to explain what is happening here?
 A: I hope you understand what they are trying to achieve: we are looking for a linear transformation $T$ which reflects points of $\Bbb R^2$ on the line $y=2x$. We will completely know a linear transformation, if we know it's action on a basis of $\Bbb R^2$. You can check independently, that the vectors
$$\begin{bmatrix}
1 \\2 
\end{bmatrix}, \begin{bmatrix}-2 \\ 1 \end{bmatrix} $$
form a basis for $\Bbb R^2. $So we will know $T$ if we know it's action on this basis. Call this basis $\beta'$. Why are we choosing this basis? This is because it's easy to see $T$'s action on $\beta'$. Since $T$ reflects points on the line $y = 2x$, you can easily check that
$$T \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix}1 \\2 \end{bmatrix}$$
this is because the point $(1,2)$ lies on the line of reflection. Similarly using geometric arguments, you can show that $T(-2,1) = (2,-1)$. Therefore the matrix of $T$ with respect to the basis $\beta'$ is $$\begin{bmatrix}1 & -2 \\2 & 1 \end{bmatrix}$$
Since we are used to dealing with vectors in the usual canonical basis, we simply do a change of basis and find the corresponding matrix of $T$ with respect to the usual basis.
