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The exercise asks us to determine whether the given orthogonal matrix represents a rotation or a reflection. If it is a rotation, give the angle of rotation; if it is a reflection, give the line of reflection.

$$ A = \begin{bmatrix} -\frac{3}{5} & -\frac{4}{5}\\[0.3em] -\frac{4}{5} & \frac{3}{5}\\[0.3em] \end{bmatrix} $$

I know you can check whether it is a reflection or rotation by calculating the determinant. So for example for the matrix above

$$det(A) = -\frac{3}{5}\cdot\frac{3}{5}-(-\frac{4}{5})\cdot(-\frac{4}{5}) = -1$$

And so that means that matrix $A$ corresponds to a reflection in $R^2$, but how do you get the line of reflection from this?

$$ B = \begin{bmatrix} -\frac{1}{2} & \frac{\sqrt{3}}{2}\\[0.3em] -\frac{\sqrt{3}}{2} & -\frac{1}{2}\\[0.3em] \end{bmatrix} $$

And for a rotation, so for example matrix $B$ given above, can you simply say that it corresponds to the rotation matrix $R$

$$ R = \begin{bmatrix} cos(\theta) & -sin(\theta)\\[0.3em] sin(\theta) & cos(\theta)\\[0.3em] \end{bmatrix} $$ So this gives

$$cos(\theta) = -\frac{1}{2}, \theta = cos^{-1}(-\frac{1}{2}) = 120^{\circ}\\ sin(\theta) = -\frac{\sqrt{3}}{2}, \theta = sin^{-1}(-\frac{\sqrt{3}}{2}) = -60^{\circ} $$

Which means that matrix $B$ corresponds to a counterclockwise rotation of $120^{\circ}$, right?

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  • $\begingroup$ Find eigen vector for eigen value q that gives reflection $\endgroup$ – Cloud JR Jun 9 at 2:22
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Here is a general guideline for $2 \times 2$ orthogonal matrices.

They have one of the two forms

$$\text{Either} \ \ R = \begin{bmatrix} a &-b\\[0.3em] b & \ \ \ a\\[0.3em] \end{bmatrix} \ \ \ \ \text{or} \ \ \ \ S = \begin{bmatrix} a & \ \ \ b\\[0.3em] b & -a\\[0.3em] \end{bmatrix}$$

with norm $1$ column vectors (thus $a^2+b^2=1$), the first case with $\det(A)=a^2+b^2=1$, the second with $\det(A)=-(a^2+b^2)=-1$.

More precisely, they have the form (you have cited the first one, the second one is less known...): $$R_{\theta} = \begin{bmatrix} \cos(\theta) & -\sin(\theta)\\[0.3em] \sin(\theta) & \ \ \ \cos(\theta)\\[0.3em] \end{bmatrix} \ \ \ \ \ \ \text{or} \ \ \ \ \ \ S_{\alpha}=\begin{bmatrix} \cos(2 \alpha) & \ \ \ \sin(2 \alpha)\\[0.3em] \sin(2 \alpha) & -\cos(2 \alpha)\\[0.3em] \end{bmatrix} $$

where $\theta$ is the rotation angle, of course, and $\alpha$ is the polar angle of the axis or symmetry i.e., the angle of one of its directing vectors with the x-axis.

Thus, for your question, once you have recognized that a matrix is a symmetry matrix, it suffices to pick the upper left coefficient $ \cos(2 \alpha)$ and identify the possible $\alpha$s, with a disambiguation brought by the knowledge of $ \sin(2 \alpha)$.

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Hint for the first question: If a vector is on the line of reflection, it is unchanged by that reflection.

For the second, $\sin \frac{2\pi}3 = \frac{\sqrt3}2$, so double-check that you’ve gotten all of the signs right.

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  • $\begingroup$ I don't understand the hint for the first question, nor do I understand what you mean with the second part. $\endgroup$ – Esoemah Mar 6 '16 at 19:35
  • $\begingroup$ @Esoemah The line of reflection is unaffected by the reflection, so look for solutions to $A\mathbf v=\mathbf v$. $\sin\frac{2\pi}3 = \sin 120° \ne -\frac{\sqrt3}2$, so the matrix that corresponds to a clockwise 120° rotation isn’t $B$. Either you’ve made a sign error along the way or gotten the direction backwards (which amounts to the same thing). $\endgroup$ – amd Mar 7 '16 at 7:40
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for the line of reflection look for a mirror lines that stays fixed under the transformation. for example, in the case of $A = \begin{bmatrix} -\frac{3}{5} & -\frac{4}{5}\\[0.3em] -\frac{4}{5} & \frac{3}{5}\\[0.3em] \end{bmatrix}$ try the line $x = \pmatrix{5\\5a}.$ we want $Ax = x$ that is $$\pmatrix{-3 -4a\\-4 + 3a } = \pmatrix{5\\5a}$$ which gives you $a= -2.$ therefore the line $y = -2x$ is the mirror.

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  • $\begingroup$ I don't really understand the last step (going from the matrix to $\theta = tan^{-1}(2)$), could you maybe expand on that a bit? $\endgroup$ – Esoemah Mar 6 '16 at 20:34
  • $\begingroup$ @Esoemah, i edited my answer. see if it makes better sense now. $\endgroup$ – abel Mar 7 '16 at 12:49

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