Having problem with Rotation and Reflection

Show the following, using matrices, combinations of linear transformations, and trigonometric identities. You must prove these in general – an example is not sufficient.

(i) The combination of a rotation and a reflection is a reflection,

(ii) Any rotation can be written as a combination of two reflections.

My Input : Considering (a,b) as a general point of reference. So (-b, a) 90 degree rotation ;(-a, -b) 180 degree rotation; (b, -a) 270 degree rotation (a, -b) reflection over x-axis; (-a, b) reflection over y-axis; (b, a) reflection over y=x; (-b, -a) reflection over y= -x

Now First let us say 90 degree rotation and then reflection over x-axis So (a,b) -> (-b,a) -> (b,-a) . This we can say as 270 degree rotation so this is not a reflection. I am confused . Please help me with detailed explanation on this . Thank you !

• If you really mean a rotation by $90^{\text o}$ and a reflection over the $x$-axis, your coordinate scheme should be $$(a,b)\mapsto(-b,a)\mapsto(-b,-a)$$ which is a reflection over $y=-x$ as you already mentioned. If instead you combine the $90^{\text o}$ rotation with a reflection over $y=-x$ you have $$(a,b)\mapsto(-b,a)\mapsto(-a,b)$$ which is then a reflection over the $y$-axis. – String Dec 9 '14 at 9:12
• Actually by proceeding in this way first we did rotation and next we did reflection such that the result is reflection of the initial value . But this being taken as our assumption of operations , how can we say that this rule is applied in a general way or Is it like its not valid in all the cases? I want to prove this with a general situation. How can this be done ? – bormtolearn Dec 9 '14 at 9:33
• I gave an answer below. – String Dec 9 '14 at 9:41

Background theory

If you draw the coordinate axes on a piece of paper and mark the base vectors $\vec e_1=\begin{pmatrix}1\\0\end{pmatrix}$ and $\vec e_2=\begin{pmatrix}0\\1\end{pmatrix}$ any rotation or reflection of that paper will be determined by the direction in which these two vectors (and thus the axes) end up pointing in. If $\vec e_1$ is transformed into $\vec v_1=\begin{pmatrix}\cos \theta\\\sin \theta\end{pmatrix}$ then we have a rotation if $\vec e_2$ is transformed into $\vec v_2=\begin{pmatrix}-\sin \theta\\\cos \theta\end{pmatrix}$ since then the axes are ordered counterclockwise like they originally were. Whereas if $\vec e_2$ ends up pointing in the direction $\vec v_2=\begin{pmatrix}\sin \theta\\-\cos \theta\end{pmatrix}$ the axes then come in clockwise order, so the "paper must have been turned upside down" which is what we call a reflection.

Now, given any point in the plane, $(a,b)$ we may write the corresponding vector as $$\vec w=\begin{pmatrix}a\\b\end{pmatrix}=a\vec e_1+b\vec e_2$$ If we apply a linear transformation to the above that takes $\vec e_1$ to $\vec v_1$ and $\vec e_2$ to $\vec v_2$ we end up with $$\vec w\mapsto a\vec v_1+b\vec v_2=\begin{pmatrix}|&&|\\v_1&&v_2\\|&&|\end{pmatrix}\begin{pmatrix}a\\b\end{pmatrix}$$ so the linear transformation is really determined by vectors $\vec v_1$ and $\vec v_2$ in the relations $\vec e_1\mapsto\vec v_1$ and $\vec e_2\mapsto\vec v_2$.

From this it follows that a rotation by the angle $\theta$ and a reflection over the line $y=\tan(\theta/2)\cdot x$ are performed by the following matrices respectively: \begin{align} rotation&=\begin{pmatrix}\cos \theta&-\sin \theta\\\sin \theta&\cos \theta\end{pmatrix}\\ &\quad\\ reflection&=\begin{pmatrix}\cos \theta&\sin \theta\\\sin \theta&-\cos \theta\end{pmatrix} \end{align}

So your job is to show that when choose angles $\theta_1$ and $\theta_2$, then
(i) If you multiply a rotation matrix using $\theta_1$ by a reflection matrix using $\theta_2$, you end up with a reflection matrix (and a third angle).
(ii) If you multiply the matrices of reflections by $\theta_1$ and $\theta_2$ you obtain a rotation, and any rotation can be obtained this way.