# T ( e 1 </msub> ) = T ( 1 , 0 ) = ( cos &#x2061;<!-- ⁡ -->

$T\left({e}_{1}\right)=T\left(1,0\right)=\left(\mathrm{cos}\theta ,\mathrm{sin}\theta \right)$
and
$T\left({e}_{2}\right)=T\left(0,1\right)=\left(-\mathrm{sin}\theta ,\mathrm{cos}\theta \right)$
and
$A=\left[T\left({e}_{1}\right)|T\left({e}_{2}\right)\right]=\left[\begin{array}{cc}\mathrm{cos}\theta & -\mathrm{sin}\theta \\ \mathrm{sin}\theta & \mathrm{cos}\theta \end{array}\right]$
When I rotate a vector $\left[\begin{array}{c}x\\ y\end{array}\right]$ I get
$\left[\begin{array}{c}{x}^{\prime }\\ {y}^{\prime }\end{array}\right]=\left[\begin{array}{c}x\cdot \mathrm{cos}\theta \phantom{\rule{thinmathspace}{0ex}}-\phantom{\rule{thinmathspace}{0ex}}y\cdot \mathrm{sin}\theta \\ x\cdot \mathrm{sin}\theta \phantom{\rule{thinmathspace}{0ex}}+\phantom{\rule{thinmathspace}{0ex}}y\cdot \mathrm{cos}\theta \end{array}\right]$
Correct me if I'm wrong, but I thought that column 1 of A $\left[\begin{array}{c}\mathrm{cos}\theta \\ \mathrm{sin}\theta \end{array}\right]$, holds the 'x' values and column 2 holds the 'y' values. What I'm confused about is why does x' contain both an x component and a y component?
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Jenna Farmer
Any vector in ${\mathbb{R}}^{2}$ can be represented as follows
$v=\left(\begin{array}{c}x\\ y\end{array}\right)=||v||\left(\begin{array}{c}\mathrm{cos}\left(\varphi \right)\\ \mathrm{sin}\left(\varphi \right)\end{array}\right).$
If
$T=\left(\begin{array}{cc}\mathrm{cos}\left(\theta \right)& -\mathrm{sin}\left(\theta \right)\\ \mathrm{sin}\left(\theta \right)& \mathrm{cos}\left(\theta \right)\end{array}\right)$
Then
$Tv=||v||\left(\begin{array}{cc}\mathrm{cos}\left(\theta \right)& -\mathrm{sin}\left(\theta \right)\\ \mathrm{sin}\left(\theta \right)& \mathrm{cos}\left(\theta \right)\end{array}\right)\left(\begin{array}{c}\mathrm{cos}\left(\varphi \right)\\ \mathrm{sin}\left(\varphi \right)\end{array}\right)=||v||\left(\begin{array}{c}\mathrm{cos}\left(\varphi \right)\mathrm{cos}\left(\theta \right)-\mathrm{sin}\left(\varphi \right)\mathrm{sin}\left(\theta \right)\\ \mathrm{cos}\left(\varphi \right)\mathrm{sin}\left(\theta \right)+\mathrm{sin}\left(\varphi \right)\mathrm{cos}\left(\theta \right)\end{array}\right)$
Now, you will recognize this as the angle sum formula. That is
$\left(\begin{array}{c}\mathrm{cos}\left(\varphi \right)\mathrm{cos}\left(\theta \right)-\mathrm{sin}\left(\varphi \right)\mathrm{sin}\left(\theta \right)\\ \mathrm{cos}\left(\varphi \right)\mathrm{sin}\left(\theta \right)+\mathrm{sin}\left(\varphi \right)\mathrm{cos}\left(\theta \right)\end{array}\right)=\left(\begin{array}{c}\mathrm{cos}\left(\varphi +\theta \right)\\ \mathrm{sin}\left(\varphi +\theta \right)\end{array}\right)$
so ultimately you can see that this matrix is simply rotating by adding the appropriate angle, and this is why you see the original components of the vector after applying the linear transformation.
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Lorena Beard
Your typical point with polar coordinates $r$ and $\varphi$ and has vector
$\left[\begin{array}{c}r\mathrm{cos}\varphi \\ r\mathrm{sin}\varphi \end{array}\right].$
Multiplying A into this gives
$\left[\begin{array}{c}r\mathrm{cos}\theta \mathrm{cos}\varphi -r\mathrm{sin}\theta \mathrm{sin}\varphi \\ r\mathrm{sin}\theta \mathrm{cos}\varphi +r\mathrm{cos}\theta \mathrm{sin}\varphi \end{array}\right]=\left[\begin{array}{c}r\mathrm{cos}\left(\theta +\varphi \right)\\ r\mathrm{sin}\left(\theta +\varphi \right)\end{array}\right].$
It's at the same distance from the origin, but rotated anticlockwise by an angle $\theta$.
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