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...
0
$\begingroup$
$\endgroup$
0
$\begingroup$
$\endgroup$
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: $$ \begin{pmatrix} x''' \\ y''' \end{pmatrix} = \frac1{\sqrt{3}} \begin{pmatrix} 1 & - \sqrt{2} \\ \sqrt{2} & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \frac1{\sqrt{3}} \begin{pmatrix} 1 & \sqrt{2} \\ -\sqrt{2} & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \\ =\frac 13 \begin{pmatrix} -1 & 2\sqrt{2} \\ 2\sqrt{2} & 1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} $$
-
-
$\begingroup$ this angle brings the x-axis to co-incide with the line $y=\sqrt{2}x$ $\endgroup$ – David Holden Nov 16 '14 at 6:39
0
$\begingroup$
$\endgroup$
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 $A$
-
-
$\begingroup$ Using coordinate geometry try to find reflection of a point $(a,b)$ through $y=mx$ $\endgroup$ – Learnmore Nov 16 '14 at 6:36
0
$\begingroup$
$\endgroup$
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.
