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In Friedberg et al. Linear Algebra there is an example that asks to find the matrix of the linear operator in the standard basis which reflects vectors of $\mathbb{R}^2$ across the line $y=2x$.

Identifying that we can reflect across any line $y=mx$ by $(x,m)\mapsto(-m,x)$ we may choose the basis $\mathcal{B'}=\{\binom{1}{2},\binom{-2}{1}\}.$

The text next claims that the operator $T$ in this basis is given by $$[T]_{\mathcal{B'}}=\begin{pmatrix}1&0\\0 &-1\end{pmatrix}.$$

However, I'm not understanding why the matrix in the basis $\mathcal{B}'$ is the matrix above. Could anyone shed some light on my point of confusion?

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    What does this reflection do to the two basis vectors? – 2017-01-15
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    The first basis vector should stay fixed (since it lies on the line) and the other should be reflected. Correct? – 2017-01-15
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    Right. Recalling that the columns of the transformation matrix are the images of the basis vectors expressed relative to the codomain (output) basis, what does that tell you about the columns of $[T]_{\mathcal{B'}}$ – 2017-01-15

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$u_1=\begin{pmatrix}{1}\\{2}\end{pmatrix}$ is a vector in the direction of the line and $u_2=\begin{pmatrix}{-2}\\{1}\end{pmatrix}$ is perpendicular, so

$$\begin{cases} T(u_1)=u_1\\T(u_2)=-u_2\end{cases}\underbrace{\Rightarrow}_{\text{transposing}\\\text{coefficients}} [T]_{\mathcal{B'}}=\begin{pmatrix}1&0\\0 &-1\end{pmatrix}.$$

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    Is $[T]_\mathcal{B'}$ the "*matrix of the linear operator in the standard basis*" that the question initially asks for? If it is then my understanding and book appears backwards in regards to what basis a matrix is in. Please explain. – 2017-01-17
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    Using $ B=S^{-1}AS$ i get $\begin{pmatrix}-3/5&4/5\\4/5 &3/5\end{pmatrix}$ Is this not T in the standard basis. Why or Why not? – 2017-01-17
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    @JPKowal 1 $(a)\;\;[T]_{\mathcal{B'}}=\begin{pmatrix}1&0\\0 &-1\end{pmatrix}$ is clealy the matrix of $T$ in the basis $B'=\{u_1,u_2\}.$ $(b)\;\;$ The matrix of $T$ in the standard basis is (as you say): $$\begin{pmatrix}1&-2\\2 &1\end{pmatrix}\begin{pmatrix}1&0\\0 &-1\end{pmatrix}\begin{pmatrix}1&-2\\2 &1\end{pmatrix}^{-1}=\dfrac{1}{5}\begin{pmatrix}-3&4\\4 & 3\end{pmatrix}.$$ – 2017-01-17