# Let vec(x) be a vector, such that vec(x) = (−1, 1), " and let " R(θ) = [(costheta, -sintheta), (sintheta, costheta )], that is Rotation Operator. For theta=3/4pi find vec(y) = R(theta)vec(x)? Make a sketch showing x, y, and θ?

Jun 2, 2016

This turns out to be a counterclockwise rotation. Can you guess by how many degrees?

Let $T : {\mathbb{R}}^{2} \mapsto {\mathbb{R}}^{2}$ be a linear transformation, where

$T \left(\vec{x}\right) = R \left(\theta\right) \vec{x} ,$
$R \left(\theta\right) = \left[\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right] ,$
$\vec{x} = \left\langle- 1 , 1\right\rangle .$

Note that this transformation was represented as the transformation matrix $R \left(\theta\right)$.

What it means is since $R$ is the rotation matrix which represents the rotational transformation, we can multiply $R$ by $\vec{x}$ to accomplish this transformation.

$\left[\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right] \times \left\langle- 1 , 1\right\rangle$

For an $M \times K$ and $K \times N$ matrix, the result is an $\textcolor{g r e e n}{M \times N}$ matrix, where $M$ is the row dimension and $N$ is the column dimension. That is:

$\left[\begin{matrix}{y}_{11} & {y}_{12} & . . . & {y}_{1 n} \\ {y}_{21} & {y}_{22} & . . . & {y}_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ {y}_{m 1} & {y}_{m 2} & . . . & {y}_{m n}\end{matrix}\right]$

$= \left[\begin{matrix}{R}_{11} & {R}_{12} & . . . & {R}_{1 k} \\ {R}_{21} & {R}_{22} & . . . & {R}_{2 k} \\ \vdots & \vdots & \ddots & \vdots \\ {R}_{m 1} & {R}_{m 2} & . . . & {R}_{m k}\end{matrix}\right] \times \left[\begin{matrix}{x}_{11} & {x}_{12} & . . . & {x}_{1 n} \\ {x}_{21} & {x}_{22} & . . . & {x}_{2 n} \\ \vdots & \vdots & \ddots & \vdots \\ {x}_{k 1} & {x}_{k 2} & . . . & {x}_{k n}\end{matrix}\right]$

Therefore, for a $2 \times 2$ matrix multiplied by a $1 \times 2$, we have to transpose the vector to get a $2 \times 1$ column vector, giving us an answer that is a $\setminus m a t h b f \left(2 \times 1\right)$ column vector.

Multiplying these two gives:

$\left[\begin{matrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta\end{matrix}\right] \times \left[\begin{matrix}- 1 \\ 1\end{matrix}\right]$

$= \left[\begin{matrix}- \cos \theta - \sin \theta \\ - \sin \theta + \cos \theta\end{matrix}\right]$

Next, we can plug in $\theta = \frac{3 \pi}{4}$ (which I'm assuming is the correct angle) to get:

$\textcolor{b l u e}{T \left(\vec{x}\right) = R \left(\theta\right) \vec{x}}$

$= R \left(\theta\right) \left[\begin{matrix}- 1 \\ 1\end{matrix}\right]$

$= \left[\begin{matrix}- \cos \left(\frac{3 \pi}{4}\right) - \sin \left(\frac{3 \pi}{4}\right) \\ - \sin \left(\frac{3 \pi}{4}\right) + \cos \left(\frac{3 \pi}{4}\right)\end{matrix}\right]$

$= \left[\begin{matrix}- \cos {135}^{\circ} - \sin {135}^{\circ} \\ - \sin {135}^{\circ} + \cos {135}^{\circ}\end{matrix}\right]$

$= \left[\begin{matrix}- \left(- \frac{\sqrt{2}}{2}\right) - \frac{\sqrt{2}}{2} \\ - \frac{\sqrt{2}}{2} + \left(- \frac{\sqrt{2}}{2}\right)\end{matrix}\right]$

$= \textcolor{b l u e}{\left[\begin{matrix}0 \\ - \sqrt{2}\end{matrix}\right]}$

Now, let's graph this to see what this looks like. I can tell that it's a counterclockwise rotation, after determining the transformed vector.

Indeed, a counterclockwise rotation by ${135}^{\circ}$.

CHALLENGE: Maybe you can consider what happens when the matrix is $\left[\begin{matrix}\cos \theta & \sin \theta \\ - \sin \theta & \cos \theta\end{matrix}\right]$ instead. Do you think it will be clockwise?