Question

How 2×2 matrix with components A B C D . Transform under So (2)

Answer 1

Answer on Question #63038 – Math – Abstract Algebra

Question

How $2 \times 2$ matrix with components A B C D. Transform under SO(2).

Solution

If $X = \left( \begin{array}{cc}A & B\\ C & D \end{array} \right), Y = \left( \begin{array}{cc}E & F\\ G & H \end{array} \right)$, where $A,B,C,D,E,F,G,H,\lambda$ are real numbers, then

$$X + Y = \left( \begin{array}{cc} A & B \\ C & D \end{array} \right) + \left( \begin{array}{cc} E & F \\ G & H \end{array} \right) = \left( \begin{array}{cc} A + E & B + F \\ C + G & D + H \end{array} \right),$$ $$X - Y = \left( \begin{array}{cc} A & B \\ C & D \end{array} \right) - \left( \begin{array}{cc} E & F \\ G & H \end{array} \right) = \left( \begin{array}{cc} A - E & B - F \\ C - G & D - H \end{array} \right),$$ $$\lambda X = \lambda \left( \begin{array}{cc} A & B \\ C & D \end{array} \right) = \left( \begin{array}{cc} \lambda A & \lambda B \\ \lambda C & \lambda D \end{array} \right),$$ $$X^{-1} = \left(\left( \begin{array}{cc} A & B \\ C & D \end{array} \right)\right)^{-1} = \frac{1}{AD - BC} \left( \begin{array}{cc} D & -B \\ -C & A \end{array} \right),$$ $$XY = \left( \begin{array}{cc} A & B \\ C & D \end{array} \right) \left( \begin{array}{cc} E & F \\ G & H \end{array} \right) = \left( \begin{array}{cc} AE + BG & AF + BH \\ CE + DG & CF + DH \end{array} \right).$$

Group SO(n) consists of $2 \times 2$ matrices satisfying conditions

$$Q^T Q = Q Q^T = I, \det(Q) = 1,$$

where elements of $Q$ are real, $Q^T$ is the transpose of $Q$ and $I$ is the identity matrix, $\det(Q)$ is the determinant of the matrix $Q$.

The group SO(2) consists of matrices of the form

$$\left( \begin{array}{cc} \cos t & -\sin t \\ \sin t & \cos t \end{array} \right),$$

where $t$ takes on real values.

If $X = \left( \begin{array}{cc}A & B\\ C & D \end{array} \right)$ and $Q = \left( \begin{array}{cc}\cos t & -\sin t\\ \sin t & \cos t \end{array} \right)$, then

$$XQ = \left( \begin{array}{cc} A & B \\ C & D \end{array} \right) \left( \begin{array}{cc} \cos t & -\sin t \\ \sin t & \cos t \end{array} \right) = \left( \begin{array}{cc} A\cos t + B\sin t & -A\sin t + B\cos t \\ C\cos t + D\sin t & -C\sin t + D\cos t \end{array} \right),$$ $$QX = \left( \begin{array}{cc} \cos t & -\sin t \\ \sin t & \cos t \end{array} \right) \left( \begin{array}{cc} A & B \\ C & D \end{array} \right) = \left( \begin{array}{cc} A\cos t - C\sin t & B\cos t - D\sin t \\ A\sin t + C\cos t & B\sin t + D\cos t \end{array} \right).$$

Other operations are performed using the general rules of matrix operations.

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