text: # 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?
  You can still ask an expert for help
  
  ## Want to know more about Matrix transformations?
  
  • Questions are typically answered in as fast as 30 minutes
  
  Solve your problem for the price of one coffee
  
  • Math expert for every subject
  • Pay only if we can solve it
  
  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
  
  Solve your problem for the price of one coffee
explanation: The document contains a detailed explanation of a matrix transformation, specifically a rotation matrix, and how it affects a vector in \(\mathbb{R}^2\). The user expresses confusion about why the transformed \(x'\) component contains both \(x\) and \(y\) components. Two responses are provided, each explaining the transformation and addressing the user's confusion.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # 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?
  You can still ask an expert for help
  
  ## Want to know more about Matrix transformations?
  
  • Questions are typically answered in as fast as 30 minutes
  
  Solve your problem for the price of one coffee
  
  • Math expert for every subject
  • Pay only if we can solve it
  
  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
  
  Solve your problem for the price of one coffee
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a detailed explanation of a matrix transformation, specifically a rotation matrix, and how it affects a vector in \(\mathbb{R}^2\). The user expresses confusion about why the transformed \(x'\) component contains both \(x\) and \(y\) components. Two responses are provided, each explaining the transformation and addressing the user's confusion.
  
  Label: question-and-answer