# 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.
###### Did you like this example?
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 $\theta$.
###### Did you like this example?

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The text passage describes a mathematical concept related to matrix transformations, specifically the rotation of a vector in a 2D space. It provides a detailed explanation of how a rotation matrix operates on a vector, and it includes a question about why the transformed x-coordinate (x') contains both x and y components from the original vector.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
# 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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## Want to know more about Matrix transformations?

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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.
###### Did you like this example?
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 $\theta$.
###### Did you like this example?

• Questions are typically answered in as fast as 30 minutes

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user
The text passage describes a mathematical concept related to matrix transformations, specifically the rotation of a vector in a 2D space. It provides a detailed explanation of how a rotation matrix operates on a vector, and it includes a question about why the transformed x-coordinate (x') contains both x and y components from the original vector.

TYPE: SINGLE_QUESTION_WITH_ANSWER