How can I represent the rotation of a point in $mathbb{R}^2$ graphically?











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Let $P = (x, y)$ be a point in the plane; now if I want to rotate it by an angle $theta$ the point $P$ will be moved from $(x, y)$ to $(X,Y)$ where:



$$X = x(cos theta) - y(sin theta)$$
$$Y = x(sin theta) + y(cos theta)$$



Why is this true, how can I represent it graphically?
Can somebody explain me how to get the representative matrix of the rotation function?



Linear transformation of R2: rotation about the origin










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  • Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
    – Dog_69
    Dec 1 at 21:08












  • @Dog_69 I don't find intuitive at all what you said there, sorry.
    – Ryk
    Dec 1 at 21:24















up vote
0
down vote

favorite












Let $P = (x, y)$ be a point in the plane; now if I want to rotate it by an angle $theta$ the point $P$ will be moved from $(x, y)$ to $(X,Y)$ where:



$$X = x(cos theta) - y(sin theta)$$
$$Y = x(sin theta) + y(cos theta)$$



Why is this true, how can I represent it graphically?
Can somebody explain me how to get the representative matrix of the rotation function?



Linear transformation of R2: rotation about the origin










share|cite|improve this question









New contributor




Ryk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
    – Dog_69
    Dec 1 at 21:08












  • @Dog_69 I don't find intuitive at all what you said there, sorry.
    – Ryk
    Dec 1 at 21:24













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Let $P = (x, y)$ be a point in the plane; now if I want to rotate it by an angle $theta$ the point $P$ will be moved from $(x, y)$ to $(X,Y)$ where:



$$X = x(cos theta) - y(sin theta)$$
$$Y = x(sin theta) + y(cos theta)$$



Why is this true, how can I represent it graphically?
Can somebody explain me how to get the representative matrix of the rotation function?



Linear transformation of R2: rotation about the origin










share|cite|improve this question









New contributor




Ryk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











Let $P = (x, y)$ be a point in the plane; now if I want to rotate it by an angle $theta$ the point $P$ will be moved from $(x, y)$ to $(X,Y)$ where:



$$X = x(cos theta) - y(sin theta)$$
$$Y = x(sin theta) + y(cos theta)$$



Why is this true, how can I represent it graphically?
Can somebody explain me how to get the representative matrix of the rotation function?



Linear transformation of R2: rotation about the origin







linear-algebra rotations






share|cite|improve this question









New contributor




Ryk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Ryk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Dec 1 at 21:15









Hans Hüttel

3,0722920




3,0722920






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asked Dec 1 at 20:29









Ryk

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163




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Ryk is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.












  • Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
    – Dog_69
    Dec 1 at 21:08












  • @Dog_69 I don't find intuitive at all what you said there, sorry.
    – Ryk
    Dec 1 at 21:24


















  • Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
    – Dog_69
    Dec 1 at 21:08












  • @Dog_69 I don't find intuitive at all what you said there, sorry.
    – Ryk
    Dec 1 at 21:24
















Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
– Dog_69
Dec 1 at 21:08






Sure. Draw a point on a paper and rotate it some angle. Then try to write the new point in terms of the old. You should be able to get it.
– Dog_69
Dec 1 at 21:08














@Dog_69 I don't find intuitive at all what you said there, sorry.
– Ryk
Dec 1 at 21:24




@Dog_69 I don't find intuitive at all what you said there, sorry.
– Ryk
Dec 1 at 21:24










2 Answers
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For any linear transformation $T: V_1 rightarrow V_2$, where $V_1$ and $V_2$ are vector spaces, its matrix representation consists of the images under $T$ of the basis vectors of $V_2$. Since $V_1 = mathbb{R}^2$, we have that the matrix for a counterclockwise rotation of $theta$ is



$$left[begin{array}{ll} cos theta & - sin theta \
sin theta & cos theta end{array}right]$$






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  • Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
    – Ryk
    Dec 1 at 21:22


















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Prove the parallelogram rule: if $(x,y)$ and $(x',y')$ are any two points, then the four points $(0,0)$, $(x,y)$, $(x',y')$ and $(x+x',y+y')$ form a parallelogram.



Now: Where does $(1,0)$ go when rotated by $theta$? Where does $(0,1)$ go? (Use the definitions of sine and cosine.)



Where does $(x,0)$ go when rotated by $theta$? Where does $(0,y)$ go? (Use the results of the last paragraph.)



Where does $(x,y)$ go when rotated by $theta$? (Hint: use the parallelogram rule on the points $(x,0)$ and $(0,y)$. Notice that when you rotate a parallelogram, you get another parallelogram. Also notice that a rectangle is a special type of parallelogram.)






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    2 Answers
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    2 Answers
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    For any linear transformation $T: V_1 rightarrow V_2$, where $V_1$ and $V_2$ are vector spaces, its matrix representation consists of the images under $T$ of the basis vectors of $V_2$. Since $V_1 = mathbb{R}^2$, we have that the matrix for a counterclockwise rotation of $theta$ is



    $$left[begin{array}{ll} cos theta & - sin theta \
    sin theta & cos theta end{array}right]$$






    share|cite|improve this answer





















    • Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
      – Ryk
      Dec 1 at 21:22















    up vote
    0
    down vote













    For any linear transformation $T: V_1 rightarrow V_2$, where $V_1$ and $V_2$ are vector spaces, its matrix representation consists of the images under $T$ of the basis vectors of $V_2$. Since $V_1 = mathbb{R}^2$, we have that the matrix for a counterclockwise rotation of $theta$ is



    $$left[begin{array}{ll} cos theta & - sin theta \
    sin theta & cos theta end{array}right]$$






    share|cite|improve this answer





















    • Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
      – Ryk
      Dec 1 at 21:22













    up vote
    0
    down vote










    up vote
    0
    down vote









    For any linear transformation $T: V_1 rightarrow V_2$, where $V_1$ and $V_2$ are vector spaces, its matrix representation consists of the images under $T$ of the basis vectors of $V_2$. Since $V_1 = mathbb{R}^2$, we have that the matrix for a counterclockwise rotation of $theta$ is



    $$left[begin{array}{ll} cos theta & - sin theta \
    sin theta & cos theta end{array}right]$$






    share|cite|improve this answer












    For any linear transformation $T: V_1 rightarrow V_2$, where $V_1$ and $V_2$ are vector spaces, its matrix representation consists of the images under $T$ of the basis vectors of $V_2$. Since $V_1 = mathbb{R}^2$, we have that the matrix for a counterclockwise rotation of $theta$ is



    $$left[begin{array}{ll} cos theta & - sin theta \
    sin theta & cos theta end{array}right]$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 1 at 21:20









    Hans Hüttel

    3,0722920




    3,0722920












    • Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
      – Ryk
      Dec 1 at 21:22


















    • Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
      – Ryk
      Dec 1 at 21:22
















    Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
    – Ryk
    Dec 1 at 21:22




    Well I know that from a booklet but how can you actually reach that matrix, what's the proof behind it?
    – Ryk
    Dec 1 at 21:22










    up vote
    0
    down vote













    Prove the parallelogram rule: if $(x,y)$ and $(x',y')$ are any two points, then the four points $(0,0)$, $(x,y)$, $(x',y')$ and $(x+x',y+y')$ form a parallelogram.



    Now: Where does $(1,0)$ go when rotated by $theta$? Where does $(0,1)$ go? (Use the definitions of sine and cosine.)



    Where does $(x,0)$ go when rotated by $theta$? Where does $(0,y)$ go? (Use the results of the last paragraph.)



    Where does $(x,y)$ go when rotated by $theta$? (Hint: use the parallelogram rule on the points $(x,0)$ and $(0,y)$. Notice that when you rotate a parallelogram, you get another parallelogram. Also notice that a rectangle is a special type of parallelogram.)






    share|cite|improve this answer

























      up vote
      0
      down vote













      Prove the parallelogram rule: if $(x,y)$ and $(x',y')$ are any two points, then the four points $(0,0)$, $(x,y)$, $(x',y')$ and $(x+x',y+y')$ form a parallelogram.



      Now: Where does $(1,0)$ go when rotated by $theta$? Where does $(0,1)$ go? (Use the definitions of sine and cosine.)



      Where does $(x,0)$ go when rotated by $theta$? Where does $(0,y)$ go? (Use the results of the last paragraph.)



      Where does $(x,y)$ go when rotated by $theta$? (Hint: use the parallelogram rule on the points $(x,0)$ and $(0,y)$. Notice that when you rotate a parallelogram, you get another parallelogram. Also notice that a rectangle is a special type of parallelogram.)






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Prove the parallelogram rule: if $(x,y)$ and $(x',y')$ are any two points, then the four points $(0,0)$, $(x,y)$, $(x',y')$ and $(x+x',y+y')$ form a parallelogram.



        Now: Where does $(1,0)$ go when rotated by $theta$? Where does $(0,1)$ go? (Use the definitions of sine and cosine.)



        Where does $(x,0)$ go when rotated by $theta$? Where does $(0,y)$ go? (Use the results of the last paragraph.)



        Where does $(x,y)$ go when rotated by $theta$? (Hint: use the parallelogram rule on the points $(x,0)$ and $(0,y)$. Notice that when you rotate a parallelogram, you get another parallelogram. Also notice that a rectangle is a special type of parallelogram.)






        share|cite|improve this answer












        Prove the parallelogram rule: if $(x,y)$ and $(x',y')$ are any two points, then the four points $(0,0)$, $(x,y)$, $(x',y')$ and $(x+x',y+y')$ form a parallelogram.



        Now: Where does $(1,0)$ go when rotated by $theta$? Where does $(0,1)$ go? (Use the definitions of sine and cosine.)



        Where does $(x,0)$ go when rotated by $theta$? Where does $(0,y)$ go? (Use the results of the last paragraph.)



        Where does $(x,y)$ go when rotated by $theta$? (Hint: use the parallelogram rule on the points $(x,0)$ and $(0,y)$. Notice that when you rotate a parallelogram, you get another parallelogram. Also notice that a rectangle is a special type of parallelogram.)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 at 21:30









        Akiva Weinberger

        13.7k12164




        13.7k12164






















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