Given two lines ${l}_{1}:y=x-3$ and ${l}_{2}:x=1$ find matrix representation of transformation $f$(in standard base) which switch lines each others and find all invariant lines of $f$.

Shea Stuart
2022-07-05
Answered

Find matrix representation of transformation

Given two lines ${l}_{1}:y=x-3$ and ${l}_{2}:x=1$ find matrix representation of transformation $f$(in standard base) which switch lines each others and find all invariant lines of $f$.

Given two lines ${l}_{1}:y=x-3$ and ${l}_{2}:x=1$ find matrix representation of transformation $f$(in standard base) which switch lines each others and find all invariant lines of $f$.

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eurgylchnj

Answered 2022-07-06
Author has **14** answers

Since the two straight lines intersect at the point $P=(1,-2)$ the transformation must have a fixed point in $P$. We can find such transformations in three steps.

1) translate the origin in $P$ with the translation ${T}_{P}^{-1}(x,y)\to (x-1,y+2)$

2) perform a simmetry $S$ with axis the strignt line passing thorough the new origin ad such that bisect the angle between the two lines.

3) return to the old origin with the translation ${T}_{P}(x,y)\to (x+1,y-2)$.

So the searched matrix has the form: $M={T}_{P}S{T}_{P}^{-1}$.

This is not a linear transformation but an affine one, and, if you want, can be represented by a $3\times 3$ matrix.

1) translate the origin in $P$ with the translation ${T}_{P}^{-1}(x,y)\to (x-1,y+2)$

2) perform a simmetry $S$ with axis the strignt line passing thorough the new origin ad such that bisect the angle between the two lines.

3) return to the old origin with the translation ${T}_{P}(x,y)\to (x+1,y-2)$.

So the searched matrix has the form: $M={T}_{P}S{T}_{P}^{-1}$.

This is not a linear transformation but an affine one, and, if you want, can be represented by a $3\times 3$ matrix.

icedagecs

Answered 2022-07-07
Author has **3** answers

I have an addition:

The translation matrices in omogeneous coordinate are:

${T}_{P}=\left[\begin{array}{ccc}1& 0& -1\\ 0& 0& 2\\ 0& 0& 1\end{array}\right]\phantom{\rule{2em}{0ex}}{T}_{P}^{-1}=\left[\begin{array}{ccc}1& 0& 1\\ 0& 0& -2\\ 0& 0& 1\end{array}\right]$

and the reflection matrix can be found noting that the angle between the bisetrix and the $x$-axis is $\theta ={\displaystyle \frac{3\pi}{8}}$, Then the matrix is:

$S=\left[\begin{array}{ccc}\mathrm{cos}2\theta & \mathrm{sin}2\theta & 0\\ \mathrm{sin}2\theta & -\mathrm{cos}2\theta & 0\\ 0& 0& 1\end{array}\right]$

Note that the invariant lines are the bisector and his orthogonal in $P$.

The translation matrices in omogeneous coordinate are:

${T}_{P}=\left[\begin{array}{ccc}1& 0& -1\\ 0& 0& 2\\ 0& 0& 1\end{array}\right]\phantom{\rule{2em}{0ex}}{T}_{P}^{-1}=\left[\begin{array}{ccc}1& 0& 1\\ 0& 0& -2\\ 0& 0& 1\end{array}\right]$

and the reflection matrix can be found noting that the angle between the bisetrix and the $x$-axis is $\theta ={\displaystyle \frac{3\pi}{8}}$, Then the matrix is:

$S=\left[\begin{array}{ccc}\mathrm{cos}2\theta & \mathrm{sin}2\theta & 0\\ \mathrm{sin}2\theta & -\mathrm{cos}2\theta & 0\\ 0& 0& 1\end{array}\right]$

Note that the invariant lines are the bisector and his orthogonal in $P$.

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