Let $L$ be a line through the origin in $\mathbb{R}^2$, and let $\alpha$ be the angle from the positive x-axis to $L$. Let $T : \mathbb{R}^2\rightarrow \mathbb{R}^2$ be the linear map given by reflecting across $L$. Use a change of basis matrix to compute the matrix of $T$ with respect to the standard basis $\beta = \{e_1,e_2\} \; \text{of} \;\mathbb{R}^2$.

So I know that my standard basis in $\mathbb{R}^2$ is $\beta= \left\{ \left( \begin{matrix} 1 \\ 0 \end{matrix} \right), \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) \right\}$ . When I visualize what the reflection is doing to these two basis vectors, I get that $$T \left[ \left( \begin{matrix} 1 \\ 0 \end{matrix} \right) \right]=\left[ \left( \begin{matrix} cos(2\alpha) \\ sin(2\alpha) \end{matrix} \right) \right]$$ and I get that $$T \left[ \left( \begin{matrix} 0 \\ 1 \end{matrix} \right) \right]=\left[ \left( \begin{matrix} sin(2\alpha) \\ -cos(2\alpha) \end{matrix} \right) \right].$$

Thus I would get:

$$\left[ \:T \:\right]_\beta=\left[ \begin{matrix} cos(2\alpha) & sin(2\alpha) \\ sin(2\alpha) &-cos(2\alpha) \end{matrix} \right]$$ as the matrix that represents $T$ in the ordered basis $\beta$.

I'm confused on what to do now? Is that all the question is asking?


You’ve gotten the right matrix, but it looks like whoever posed this problem intended for you to solve it a different way.

If you decompose an arbitrary vector into its components parallel to and orthogonal to the line, the reflection $L$ can be characterized as reversing the orthogonal component. In a suitable basis, then, the matrix of $L$ is simply $\operatorname{diag}(1,-1)$. Your task in this exercise is to find such a basis and apply a change-of-basis operation to this matrix to get the matrix relative to the standard basis.


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