Prove a matrix belonging to a group represents a rotation

[SNOOTCHIEBOOCHEE]

Homework Statement

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

Homework Equations

In case you are used to different notation, SO₂= {A$$\in$$ GL_n(R)| A^tA=I, Det A=1}

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 $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$ * $$\left(\begin{array}{cc}a&c\\b&d\end{array}\right)$$= [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 $$\left(\begin{array}{cc}a&b\\c&d\end{array}\right)$$ 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 I am think i did this second part wrong.

 

[Mark44]

Homework Statement

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

Homework Equations

In case you are used to different notation, SO₂= {A$$\in$$ GL_n(R)| A^tA=I, Det A=1}

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

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

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 I am 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.