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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?

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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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