Did I Make a Mistake in Finding the Determinant of an Orthogonal 2x2 Matrix?

[ognik]

Find the general form of an orthogonal 2 x 2 matrix = $ \begin{bmatrix}a&b\\c&d\end{bmatrix}$

I used $det(A)=\pm 1$ (from an earlier exercise) - and the special form of $ A_{2 \times 2}^{-1} = \frac{ \begin{bmatrix}d&-b\\-c&a\end{bmatrix}}{|A|} $ using $|A| = +1$ first, to get an 'alternative' $ A = \left(A^{-1}\right)^T = \begin{bmatrix}d&-c\\-b&a\end{bmatrix}$

This gave me $ a=Cos\theta, b=Sin \theta $ as expected so far.

But if I do something similar for c and d I get $ c=Cos\theta, d=Sin \theta $, which can't be right 'cos that would be singular.

So using |A| = -1, I get a 2nd 'alternate' $A = -\begin{bmatrix}d&-c\\-b&a\end{bmatrix} = \begin{bmatrix}-d&c\\-b&a\end{bmatrix} $ This is clearly wrong, but can't see what I've done wrong?

 

[Deveno]

Uhh...$-\begin{bmatrix}d&-c\\-b&a\end{bmatrix} = \begin{bmatrix}-d&c\\b&-a\end{bmatrix}$

 

[topsquark]

Find the general form of an orthogonal 2 x 2 matrix = $ \begin{bmatrix}a&b\\c&d\end{bmatrix}$

I used $det(A)=\pm 1$ (from an earlier exercise) - and the special form of $ A_{2 \times 2}^{-1} = \frac{ \begin{bmatrix}d&-b\\-c&a\end{bmatrix}}{|A|} $ using $|A| = +1$ first, to get an 'alternative' $ A = \left(A^{-1}\right)^T = \begin{bmatrix}d&-c\\-b&a\end{bmatrix}$

Stop it here.
[math]A = \left ( A^{-1} \right )^T[/math]

[math]\left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) = \left ( \begin{matrix} d & -c \\ -b & a \end{matrix} \right )[/math]

So by comparison we know that d = a, -c = b, -b = c, a = d. And |A| = 1. That will give you the rest.

-Dan

 

[ognik]

Uhh...$-\begin{bmatrix}d&-c\\-b&a\end{bmatrix} = \begin{bmatrix}-d&c\\b&-a\end{bmatrix}$

You mean Duh - I was stuck in Det mode :-)

 

[ognik]

Stop it here.
[math]A = \left ( A^{-1} \right )^T[/math]

[math]\left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) = \left ( \begin{matrix} d & -c \\ -b & a \end{matrix} \right )[/math]

So by comparison we know that d = a, -c = b, -b = c, a = d. And |A| = 1. That will give you the rest.

-Dan

That gave me $ Det\left ( \begin{matrix} d & -c \\ c & d \end{matrix} \right ) = d^2 + c^2 = 1, ie \:c=Sin \theta etc \:$
But I know the 2nd row of A should be $c=-Sin \theta, d= Cos \theta $

 

[topsquark]

Stop it here.
[math]A = \left ( A^{-1} \right )^T[/math]

[math]\left ( \begin{matrix} a & b \\ c & d \end{matrix} \right ) = \left ( \begin{matrix} d & -c \\ -b & a \end{matrix} \right )[/math]

So by comparison we know that d = a, -c = b, -b = c, a = d. And |A| = 1. That will give you the rest.

-Dan

That gave me $ Det\left ( \begin{matrix} d & -c \\ c & d \end{matrix} \right ) = d^2 + c^2 = 1, ie \:c=Sin \theta etc \:$
But I know the 2nd row of A should be $c=-Sin \theta, d= Cos \theta $

d = a, -c = b implies [math]A = \left ( \begin{matrix} a & b \\ -b & a \end{matrix} \right )[/math] and |A| = 1.

From the determinant we get [math]|A| = a^2 + b^2 = 1[/math].

The solutions for A are:
[math]\left ( \begin{matrix} \pm sin( \theta ) & \pm cos( \theta ) \\ \mp cos(\theta) & \pm sin(\theta) \end{matrix} \right )[/math]

and
[math]\left ( \begin{matrix} \pm cos( \theta ) & \pm sin( \theta ) \\ \mp sin(\theta) & \pm cos(\theta) \end{matrix} \right )[/math]
where the choice of [math]\pm[/math] on the sines is independent of the [math]\pm[/math] on the cosines.

-Dan

 

[ognik]

that makes sense of course, so I wonder why my book said to compare the answer with the 2-D rotation matrix?

So I was looking for something that would give that...

 

[topsquark]

that makes sense of course, so I wonder why my book said to compare the answer with the 2-D rotation matrix?

So I was looking for something that would give that...

Given any number of definitions of your x and y axes they are all rotation matrices. Note that
[math]\left ( \begin{matrix} cos( \theta) & -sin( \theta ) \\ sin( \theta) & cos(\theta) \end{matrix} \right )[/math]
is the "usual" rotation matrix.

-Dan

 

[ognik]

I think I found what I was looking for... $A \times A^T$ = identity matrix. Solving gives 2 equations - we already knew $ a^2 + b^2 =1 $, but we also get ac + bd = 0, ac = -bd which tells me that if a and b are positive one of c or d must be negative. This way there is also no doubt about using $A^{-1}$ as 1 or -1 and its a simpler solution (assuming I got it right)