# Matrix with linear transformation with reflection

Find the matrix of the linear transformation A which is the reflection in the line $y = \sqrt{2}x$ with respect to the standard basis in $\mathbb{R^2}$.

I Have no idea how to approach this problem...

## 3 Answers

rotate the co-ordinate system anti-clockwise through $\arctan \sqrt{2}$ $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \frac1{\sqrt{3}} \begin{pmatrix} 1 & \sqrt{2} \\ -\sqrt{2} & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$ now reflect in $x'$-axis $$\begin{pmatrix} x'' \\ y'' \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} x' \\ y' \end{pmatrix}$$ finally, return to original coordinates: $$\begin{pmatrix} x''' \\ y''' \end{pmatrix} = \frac1{\sqrt{3}} \begin{pmatrix} 1 & - \sqrt{2} \\ \sqrt{2} & 1 \end{pmatrix} \begin{pmatrix} x'' \\ y'' \end{pmatrix}$$ altogether: $$\begin{pmatrix} x''' \\ y''' \end{pmatrix} = \frac1{\sqrt{3}} \begin{pmatrix} 1 & - \sqrt{2} \\ \sqrt{2} & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \frac1{\sqrt{3}} \begin{pmatrix} 1 & \sqrt{2} \\ -\sqrt{2} & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ =\frac 13 \begin{pmatrix} -1 & 2\sqrt{2} \\ 2\sqrt{2} & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$$

• why did you do arctan to rotate it? – user192898 Nov 16 '14 at 6:37
• this angle brings the x-axis to co-incide with the line $y=\sqrt{2}x$ – David Holden Nov 16 '14 at 6:39

The image of $(1,0)$ when reflected through the given line becomes $(-1/3,2\sqrt{2}/3)$ and for $(0,1)$ it becomes $(2\sqrt{2}/3,1/3)$ Take them as the columns of $A$

• how did you get to this answer. – user192898 Nov 16 '14 at 6:24
• Using coordinate geometry try to find reflection of a point $(a,b)$ through $y=mx$ – Learnmore Nov 16 '14 at 6:36

Rather than give you the answer, I'll show you how to do this kind of problem in general. You have an easy description (and matrix) of your linear transformation on a specially chosen basis; transform that to standard.

Let $\mathcal B=[b_1,b_2]$ be a basis, where $b_1$ lies on that line and $b_2$ is orthogonal to it. Then on that basis the matrix of your linear transformation will be $$A'=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$$ Now you need to do bas change to the standard basis, which is given by $A=BA'B^{-1}$ where $B$ is the matrix with as columns the coordinates of $b_1,b_2$ in the standard basis.

Multiplication by $B$ converts coordinates in the basis $\mathcal B$ into the coordinates of the same vector in the standard basis; therefore what $A=BA'B^{-1}v$ is doing is convert the standard coordinates of $v$ into its coordinates on$~\mathcal B$, then compute the linear transformation in those coordinates, which is easy ($A'$), and finally convert the result back to standard coordinates.

Note that you can scale $b_1,b_2$ freely, as long as the lie on the indicated lines. So you may choose $[b_1,b_2]$ to be orthonormal, which simplifies the computation of$~B^{-1}$ since it will be the transpose of$~B$; however this might give uglier entries of$~B$ to begin with. Usually I try to stick with rational coordinates if possible, but here the very definition of the line to reflect in makes it impossible. You should try what scaling works best.