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Prove a matrix belonging to a group represents a rotation

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove algebraically that a real 2x2 matrix [tex]\left(\begin{array}{cc}a&b\\c&d\end{array}\right)[/tex] represents a rotaion iff it is in SO2

    2. Relevant equations

    In case you are used to different notation, SO2= {A[tex]\in[/tex] GLn(R)| AtA=I, Det A=1}

    3. The attempt at a solution

    ok since this is an iff statement, we have to show both directions.

    Case1: if the matrix is in SO2 then it represents a rotation

    so we know that [tex]\left(\begin{array}{cc}a&b\\c&d\end{array}\right)[/tex] * [tex]\left(\begin{array}{cc}a&c\\b&d\end{array}\right)[/tex]= [tex]\left(\begin{array}{cc}1&0\\0&1\end{array}\right)[/tex]

    also ad-bc=1

    also if its helpful [tex]\left(\begin{array}{cc}a&b\\c&d\end{array}\right)[/tex] * [tex]\left(\begin{array}{cc}a&c\\b&d\end{array}\right)[/tex]= [tex]\left(\begin{array}{cc}a2+b2&ac+bd\\ca+db&b2+d2\end{array}\right)[/tex]

    I know i can set this equal to the identity and probably solve for some stuff. but how exactly do i prove that it is a roation? moreover i am completley lost in the other direction.

    Edit: maybe i figured it out??!?

    Other direction: If [tex]\left(\begin{array}{cc}a&b\\c&d\end{array}\right)[/tex] is a rotation, then it is in SO2

    every rotaion through an angle theta can be written as cos -sin sin cos. just show that its transpose * it =1? and that its det 1? which is basically trivial... so im think i did this second part wrong.
  2. jcsd
  3. Nov 17, 2008 #2


    Staff: Mentor

    The change in notation wasn't much help, since there are now two symbols I don't know:
    [tex]SO_2 \; and GL_n(R)[/tex]

    The entry in the lower right corner of the matrix above should be c^2 + d^2.
    From your multication of AA^t, you know that a^2 + b^2 = 1, c^2 + d^2 = 1, and ac + bd = 0. A^t * A should also be I, and you get a^2 + c^2 = 1, b^2 + d^2 = 1, and ab + cd = 0.

    Let's look at Av, where v = (x, y)
    Av = (ax + by, cx + dy)
    Now, let's calculate the magnitude of Av
    [tex]|Av|^2 = |(ax + by, cx + dy)|^2
    = (ax + by)^2 + (cx + dy)^2[/tex]
    [tex]= a^2 x^2 + 2abxy + b^2y^2 + c^2x^2 + 2cdxy + d^2y^2[/tex]
    [tex]= (a^2 + c^2)x^2 + (b^2 + d^2)y^2 + 2(ab + cd)xy[/tex]
    [tex]= x^2 + y^2
    = |v|[/tex]
    The simplification down to x^2 + y^2 occurs because of what we know about A^t * A = I.
    Now, since |Av| = |v|, multiplication by A doesn't change the length of v, so maybe we can conclude that it constitutes a rotation of some sort.
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