Prove a matrix belonging to a group represents a rotation

[SNOOTCHIEBOOCHEE]

1. The problem statement, all variables and given/known data

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

2. Relevant equations

In case you are used to different notation, SO2= {A\in 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 \left(\begin{array}{cc}a&b\\c&d\end{array}\right) * \left(\begin{array}{cc}a&c\\b&d\end{array}\right)= \left(\begin{array}{cc}1&0\\0&1\end{array}\right)

also ad-bc=1

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

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 \left(\begin{array}{cc}a&b\\c&d\end{array}\right) 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.

[Mark44]

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

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



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 \left(\begin{array}{cc}a&b\\c&d\end{array}\right) * \left(\begin{array}{cc}a&c\\b&d\end{array}\right)= \left(\begin{array}{cc}1&0\\0&1\end{array}\right)

also ad-bc=1

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

The entry in the lower right corner of the matrix above should be c^2 + d^2.



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 \left(\begin{array}{cc}a&b\\c&d\end{array}\right) 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.

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
|Av|^2 = |(ax + by, cx + dy)|^2
= (ax + by)^2 + (cx + dy)^2
= a^2 x^2 + 2abxy + b^2y^2 + c^2x^2 + 2cdxy + d^2y^2
= (a^2 + c^2)x^2 + (b^2 + d^2)y^2 + 2(ab + cd)xy
= x^2 + y^2
= |v|
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.