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EDIT:

The following edit is a proposed solution to the problem, based on feedback given in the comments below.

Let $a_1, a_2$ and $a_3$ be the vectors $(1,0,0), (0,1,0)$ and $(0,0,1)$ respectively (the standard basis vectors of $\mathbb{R}^3$).

By considering Gram-Schmidt orthogonlization, @Michael has proposed the following method for finding the reflection of each basis vector in the plane $v^{\perp}$

\begin{align*} s_v(a)=a-2\frac{\langle a,v\rangle}{\|v\|^2}v. \end{align*}

Calculating this for $s_v(a_{1-3})$ yields the following vectors (In partially simplified form)

$ (\frac{5\sqrt{2} - 18}{5\sqrt{2}}, -\frac{24}{5\sqrt{2}}, -\frac{6}{\sqrt{2}}), \qquad (-\frac{24}{5\sqrt{2}}, \frac{5\sqrt{2} - 32}{5\sqrt{2}}, -\frac{8}{\sqrt{2}}), \qquad (-\frac{6}{\sqrt{2}}, -\frac{8}{\sqrt{2}}, -\frac{\sqrt{2} - 10}{\sqrt{2}}). $

The transformation matrix $M$ is composed by combining the above vectors. Thus,

$$ M = \begin{bmatrix} \frac{5\sqrt{2} - 18}{5\sqrt{2}} & -\frac{24}{5\sqrt{2}} & -\frac{6}{\sqrt{2}} \\ -\frac{24}{5\sqrt{2}} & \frac{5\sqrt{2} - 32}{5\sqrt{2}} & -\frac{8}{\sqrt{2}} \\ -\frac{6}{\sqrt{2}} & -\frac{8}{\sqrt{2}} & -\frac{\sqrt{2} - 10}{\sqrt{2}} \end{bmatrix} $$

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I am a beginner. Because of deficits in my understanding, I am having difficulty with the following question.

"Let $Sv: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be the reflection in the plane $v^{\perp} \subset \mathbb{R}^3$ with $v = (3,4,5)^t$. provide the matrix representation $M = M^{E}_{E}(s_v)$ of $s_v$ with respect to the standard basis of $\mathbb{R}^3$."

I think I have worked out the broad aspects of an approach. I would specifically appreciate help if someone could

a) Confirm that my idea is correct b) Provide assistance with the individual steps.

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Progress:

  • I dont really understand the notation $\textit{"$M = M^{E}_{E}(s_v)$"}$. Is this simply referring to the transformation matrix?

  • I dont know how to calculate the matrix representation for a linear transformation, however know that if a mapping takes one vector ($v_1$) and produces another in a perpendicular plane ($v_2$), this would mean that $v_2 \cdot v_1 = 0$. Is this a step in the right direction?

  • For the condition $\textit{"with respect to the standard basis of $\mathbb{R}^3$."}$, It seems that this answer gives steps on how to do this- so If i understand correctly, once I find a transformation matrix, I need to find another two Matrices $C^{-1}$ & $C$. I just don't understand how to calculate them.

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  • $\begingroup$ First and third point: it means the matrix of the endormorphism $S_v$ expressed in the canonical basis $E$. Second point: this might be helpful. $\endgroup$ – anderstood Jan 20 '18 at 14:18
  • $\begingroup$ You forgot to specify $v$ ... $\endgroup$ – Michael Hoppe Jan 23 '18 at 21:37
  • $\begingroup$ $s_v(1,0,0)=(0.64,-0.48,-0.6)$. $\endgroup$ – Michael Hoppe Jan 23 '18 at 21:43
  • $\begingroup$ The reflection formula is not essentially related to Gram–Schmidt. $\endgroup$ – Michael Hoppe Jan 23 '18 at 21:48
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Obviously $$s_v(a)=a-2\frac{\langle a,v\rangle}{\|v\|^2}v.$$ Now the column vectors of the matrix of a linear map with respect to the standard basis are their images under that map. So the first column would be $s_v((1,0,0))$.

EDIT (regarding the obviouslibility):

Let $v$ be the plane's normal, then $\frac{v}{\|v\|}$ is a unit normal vector. Hence $\langle a,\frac{v}{\|v\|} \rangle$ is the component of $a$ in direction $v$, so $\langle a,\frac{v}{\|v\|} \rangle\frac{v}{\|v\|}=\frac{\langle a,v\rangle }{\|v\|^2}v$ is the projection of $a$ along $v$. To get the reflected vector on that plane, just subtract the double of the projection from $a$ yielding in $a-2\frac{\langle a,v\rangle }{\|v\|^2}v$.

EDIT$^2$ (regarding $\textit{"$M = M^{E}_{E}(s_v)$"}$)

The column vectors of the representation matrix in respect to the basis $E$ are the images of those vectors unter $s_v$. Maybe this post If $T$ is a linear map and $\mathcal M(T)$ is its matrix, what exactly does the multiplication $\mathcal M(T)v$ mean? clarifies.

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  • $\begingroup$ im sorry, but this isn't obvious to me.. Perhaps you could provide some clarification? Thanks for your help $\endgroup$ – Oscar Jan 22 '18 at 19:42
  • $\begingroup$ Thanks for the extra information. I have recently covered Gram-Schmidt Orthonormalization, and it seems your answer is quite similiar to that. So in order to get my transformation matrix, is it correct that I need to repeat the above method of projection for the remaining standard basis vectors (0,1,0) and (0,0,1) ? $\endgroup$ – Oscar Jan 23 '18 at 16:45
  • $\begingroup$ @Oscar Exactly, that's what it's all about. $\endgroup$ – Michael Hoppe Jan 23 '18 at 16:50
  • $\begingroup$ Okay, to check I have understood the steps, I will calculate the transformation matrix using this method. Update to come shortly. $\endgroup$ – Oscar Jan 23 '18 at 16:55
  • $\begingroup$ I have edited the question based on your feedback. If you believe my interpretation is correct, then let me know and i will be happy to accept your answer as the solution. $\endgroup$ – Oscar Jan 23 '18 at 19:08

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