0
$\begingroup$

Let $\bf{A}$ be a linear transformation in a 3D vector space that represents a reflection in the plane $$x_1 \sin \theta =x_2 \cos \theta$$ Find the matrix that represents this linear transformation


As the question didn't specify the basis to be standard basis, I am thinking of a basis on that plane and apply two points through that transformation. But I'm unfamiliar to the equation that represents the plane, usually it's the form of $\bf{n}.(\bf{x-x_0})=0$, and I don't know where to start with.

$\endgroup$
1
  • $\begingroup$ Usually, when the basis is unspecified, the standard basis is implied. $\endgroup$ – amd Dec 29 '17 at 1:01
1
$\begingroup$

The equation of the plane is:

$$x_1 \sin \theta -x_2 \cos \theta=0$$

To write the matrix for the reflection it is convenient at first assume as basis the normal vector to the plane and two linearly independent vectors in the plane:

$v_1=(\sin \theta,-\cos \theta,0)$

$v_2=(\cos \theta,\sin \theta,0)$

$v_3=(0,0,1)$

with respect to this basis to the canonical the reflection matrix is

$$A=\left( \begin{array}{cc} -\sin \theta & \cos \theta & 0 \\ \cos \theta & \sin \theta & 0 \\ 0& 0 & 1 \end{array} \right)$$

to find the matrix with respect to the canonical basis you need a change of basis.

For the change of basis let's consider the matrix $M$ which columns are the basis vectors $v_i$ which components are expressed with respect to the canonical basis:

$$M=\left( \begin{array}{cc} \sin \theta & \cos \theta & 0 \\ -\cos \theta & \sin \theta & 0 \\ 0& 0 & 1 \end{array} \right)$$

thus any vector $v$, given with respect to the $v_i$ basis, can be expressed with respect to the canonical basis as follow:

$$w=Mv\implies v=M^{-1}w$$

Finally, since the reflection from the $v_i$ basis to the canonical is given by:

$$u=Av$$

in the canonical basis the reflection is given by:

$$u=AM^{-1}w$$

$\endgroup$
3
  • $\begingroup$ @amd ops...you're absolutely right, thanks! $\endgroup$ – user Dec 29 '17 at 1:03
  • $\begingroup$ Well, w/r to the chosen basis, the reflection matrix is actually $\operatorname{diag}(-1,1,1)$. Your matrix $A$ has as its “input” basis the special basis, but it outputs vectors with coordinates relative to the standard basis, which is not what’s usually meant by a matrix w/r to a (single) basis. $\endgroup$ – amd Dec 29 '17 at 1:36
  • $\begingroup$ @amd yes you're right, I've explained after that the matrix A is from basis $v_i$ to the canonical. $\endgroup$ – user Dec 29 '17 at 1:40
1
$\begingroup$

Your plane can be written in the form

$$\begin{pmatrix}\sin\theta\\-\cos\theta\\0\end{pmatrix}\cdot\begin{pmatrix}x_{1}\\x_{2}\\x_{3}\\\end{pmatrix}=0$$

which is the usual form with $\boldsymbol{x}_{0}=0$ and

$$\boldsymbol{n}=\begin{pmatrix}\sin\theta\\-\cos\theta\\0\end{pmatrix}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.