Definition \(\textrm{exp}(M) = I + \sum_{k=1}^{\infty} \frac{1}{k!} M^k\) ; M is a square matrix. Case I \(M = \begin{pmatrix} d_1 & 0 & 0 & 0 & 0 & \dots & 0 \\ 0 & d_2 & 0 & 0 & 0 & \dots & 0 \\ 0 & 0 & d_3 & 0 & 0 & \dots & 0 \\ 0 & 0 & 0 & d_4 & 0 & \dots & 0 \\ 0 & 0 & 0 & 0 & d_ 5 & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \dots & d_n \end{pmatrix} \\ \textrm{exp}(M) = \begin{pmatrix} e^{d_1} & 0 & 0 & 0 & 0 & \dots & 0 \\ 0 & e^{d_2} & 0 & 0 & 0 & \dots & 0 \\ 0 & 0 & e^{d_3} & 0 & 0 & \dots & 0 \\ 0 & 0 & 0 &e^{ d_4} & 0 & \dots & 0 \\ 0 & 0 & 0 & 0 & e^{d_ 5} & \dots & 0 \\ \vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & 0 & 0 & \dots & e^{d_n} \end{pmatrix} \) Corollary: M = k I exp(M) = exp(k) I Case II \(M = S J S^{-1} \\ \textrm{exp}(M) = S \textrm{exp}(J) S^{-1}\) Case III M = (A + B) exp(M) = exp(A) exp(B) if A B = B A Case IV \(M = a \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ \textrm{exp}(M) = \begin{pmatrix} \cosh(a) & \sinh(a) \\ \sinh(a) & \cosh(a) \end{pmatrix}\) Case IV \(M = a \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \\ \textrm{exp}(M) = \begin{pmatrix} \cos(a) & - \sin(a) \\ \sin(a) & \cos(a) \end{pmatrix}\) Case V \(M = \begin{pmatrix} 0 & a \\ b & 0 \end{pmatrix} \\ \textrm{exp}(M) = \begin{pmatrix} \cosh \left( \sqrt{a} \sqrt{b} \right) & \frac{\sqrt{a}}{\sqrt{b}} \sinh \left( \sqrt{a} \sqrt{b} \right) \\ \frac{\sqrt{b}}{\sqrt{a}} \sinh \left( \sqrt{a} \sqrt{b} \right) & \cosh \left( \sqrt{a} \sqrt{b} \right) \end{pmatrix} ; \quad a \neq 0 \neq b \) Case VI \(M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ \textrm{exp}(M) = \begin{pmatrix} e^a & b e^{\frac{a+d}{2}} \frac{ 2 \sinh \frac{a - d}{2} }{ a-d} \\ c e^{\frac{a+d}{2}} \frac{ 2 \sinh \frac{a - d}{2} }{ a-d} & e^ b \end{pmatrix} = \begin{pmatrix} e^a & b \frac{ e^a - e^d }{ a-d} \\ c \frac{ e^a - e^d }{ a-d} & e^ b \end{pmatrix} ; \quad b c = 0, a\neq d \) Case VII \(M = \begin{pmatrix} a & b \\ c & a \end{pmatrix} \\ \textrm{exp}(M) = \begin{pmatrix} e^a & b e^a \\ c e^a & e^a \end{pmatrix} ; \quad b c = 0 \) Case VIII \(M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ \textrm{exp}(M) = \frac{e^{\frac{a+d}{2}}}{\sqrt{(a-d)^2+4bc}} \begin{pmatrix} \sqrt{(a-d)^2+4bc} \cosh \frac{\sqrt{(a-d)^2+4bc}}{2} + (a-d) \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} & 2 b \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} \\ 2 c \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} & \sqrt{(a-d)^2+4bc} \cosh \frac{\sqrt{(a-d)^2+4bc}}{2} + (d-a) \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} \end{pmatrix} \)
A happy chance. I have found 'learning matrices' to be deadly boring and have succumbed to motivational failure on several occasions. Spice up matrices with exp and it becomes irresistible - many thanks (not for the first time). -C2.
Me, too. I got far enough into it to get a nice headache. I'll try more tomorrow, God willin' and the creek don't rise. Thank you! Please Register or Log in to view the hidden image!
Whereas the Taylor expansion of exp(X) is \( 1 + \sum_{k=1}^{\infty} \frac{1}{k!} X^k\), but not for any X? Since if X is a matrix it must be a square matrix?
If X is a real or complex number, it works for any X. The matrix product AB is only defined if A is a \(n \times m\) matrix and B is a \(m \times p \) matrix because each element of the product is the dot product of a row of A with a column of B. \( \left( AB \right)_{ij} = \sum_{k=1}^{m} A_{ik} \times B_{kj}\). As a result BA might not be possible even if AB is and \(M^2\) is only defined for square matrixes.
If we take your cases IV and V, i.e. let \( A = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\;\; B = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} \), then AB = -BA. What happens then [ed: well, I can see that A + B will square to zero]? Or suppose \( B = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \), then \( A + B = \begin{pmatrix} 0 & 1 - i \\ 1 + i & 0 \end{pmatrix} \).
\(M=\begin{pmatrix} 0 & 1 - i \\ 1 + i & 0 \end{pmatrix} = \begin{pmatrix} - \frac{1 -i}{2} & \frac{1 -i}{2} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} - \sqrt{2} & 0 \\ 0 & \sqrt{2} \end{pmatrix} \begin{pmatrix} - \frac{1 +i}{2} & \frac{1}{\sqrt{2}} \\ \frac{1 +i}{2} & \frac{1}{\sqrt{2}} \end{pmatrix} \) So we have: \(M^{2n} = 2^{n} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = 2^{n} I \\ M^{2n+1} = 2^{n} \begin{pmatrix} 0 & 1 - i \\ 1 + i & 0 \end{pmatrix} = 2^{n} M\) So \(\textrm{exp}(M) = \sum_{k=0}^{\infty} \left( \frac{ 2^{k} }{ (2k)! } I +\frac{ 2^{k} }{ (2k+1)! } M \right) = \left( \cosh \sqrt{2} \right) I + \left( \frac{1}{\sqrt{2}} \sinh \sqrt{2} \right) M \\ \quad \quad = \begin{pmatrix} \cosh \sqrt{2} & \frac{1 - i}{\sqrt{2}} \sinh \sqrt{2} \\ \frac{1 + i}{\sqrt{2}} \sinh \sqrt{2} & \cosh \sqrt{2} \end{pmatrix}\) Alternately: \(\textrm{exp}(M) = \begin{pmatrix} - \frac{1 -i}{2} & \frac{1 -i}{2} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} e^{- \sqrt{2}} & 0 \\ 0 & e^{\sqrt{2}} \end{pmatrix} \begin{pmatrix} - \frac{1 +i}{2} & \frac{1}{\sqrt{2}} \\ \frac{1 +i}{2} & \frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} \cosh \sqrt{2} & \frac{1 - i}{\sqrt{2}} \sinh \sqrt{2} \\ \frac{1 + i}{\sqrt{2}} \sinh \sqrt{2} & \cosh \sqrt{2} \end{pmatrix}\)
No, they just need to be square matrices of the same size. However if A and B don't commute, the result is not I. To second order, (I+A+A^2/2)(I+B+B^2/2)(I-A+A^2/2)(I-B+B^2/2) = (I+A+B+AB+A^2/2+B^2/2+...) (I-A-B+AB+A^2/2+B^2/2+...) = I+AB-BA+... But if A is Hermitian, so is exp(A). And if A and B are Hermitian and commute, then AB is Hermitian.
Ok. I have a formula that expressly uses two of the Pauli matrices: \( \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\;\; \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \). But you include an angle, \( \theta \), as: \( e^{-i\theta \sigma_y}e^{-i\theta \sigma_x}e^{i\theta \sigma_y}e^{i\theta \sigma_x}\). Then if \( \theta \) is small, higher order terms are small, so you get good approximation to order 3 in the expansion. The "trick" is that rotations about x and y (as axes in Euclidean 3-space), also generate rotations about the z axis, thanks to the equality: \( \sigma_x \sigma_y = i \sigma_z \).
As from post #10, we have \(\lim \limits_{\theta \to 0} e^{-i\theta \sigma_y}e^{-i\theta \sigma_x}e^{i\theta \sigma_y}e^{i\theta \sigma_x} = e^{\theta^2 ( \sigma_x \sigma_y - \sigma_y \sigma_x ) } = e^{\theta^2 [ \sigma_x , \sigma_y ] } = e^{2 i \theta^2 \sigma_z} \) Analytically, we have for \( \sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} , \quad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}\), \(e^{i \theta \sigma_x} = (\cos \theta) I + i (\sin \theta) \sigma_x \\ e^{i \theta \sigma_y} = (\cos \theta) I + i (\sin \theta) \sigma_y \\ e^{i \theta \sigma_z} = (\cos \theta) I + i (\sin \theta) \sigma_z \\ e^{-i\theta \sigma_y}e^{-i\theta \sigma_x}e^{i\theta \sigma_y}e^{i\theta \sigma_x} = ( \cos 2\theta + \frac{1}{2} \sin^2 2 \theta ) I + \frac{i}{2} (1 - \cos 2 \theta) (\sin 2 \theta) ( \sigma_x - \sigma_y) + \frac{i}{2} (\sin^2 2\theta) \sigma_z\) \(\lim \limits_{\theta \to 0} e^{-i\theta \sigma_y}e^{-i\theta \sigma_x}e^{i\theta \sigma_y}e^{i\theta \sigma_x} \\ \quad = \left(1-2\theta^4+\frac{4}{3} \theta^6 \right) I + i \left( 2 \theta^3 - 2 \theta^5 + \frac{4}{5} \theta^7 \right) ( \sigma_x - \sigma_y) + i \left ( 2 \theta^2- \frac{8}{3} \theta^4 + \frac{64}{45} \theta^6 \right) \sigma_z \) \(\begin{array}{c|ccc} \quad & \times \sigma_x & \times \sigma_y & \times \sigma_z \\ \hline \\ \sigma_x & I & i \sigma_z & - i \sigma_y \\ \sigma_y & -i \sigma_z & I & i \sigma_x \\ \sigma_z & i \sigma_y & -i \sigma_x & I \end{array}\) // \(e^{i \theta_x \sigma_x + i \theta_y \sigma_y + i \theta_z \sigma_z} = \left( \cos \sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2} \right) I + i \frac{\theta_x}{\sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2}} \left( \sin \sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2} \right) \sigma_z + i \frac{\theta_y}{\sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2}} \left( \sin \sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2} \right) \sigma_y + i \frac{\theta_z}{\sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2}} \left( \sin \sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2} \right) \sigma_z\)
Some special cases I worked out long-hand for another thread: \(\vec{\zeta} \to A(\vec{\zeta}) = \begin{pmatrix}0 & \zeta_x & \zeta_y & \zeta_z \\ \zeta_x & 0 & 0 & 0 \\ \zeta_y & 0 & 0 & 0 \\ \zeta_z & 0 & 0 & 0 \end{pmatrix} \\ A(\vec{\zeta})^2 = \begin{pmatrix}\zeta^2 & 0 & 0 & 0 \\ 0 & \zeta_x^2 & \zeta_x \zeta_y & \zeta_x \zeta_z \\ 0 & \zeta_x \zeta_y & \zeta_y^2 & \zeta_y \zeta_z \\ 0 & \zeta_x \zeta_z & \zeta_y \zeta_z & \zeta_z^2 \end{pmatrix} \\ A(\vec{\zeta})^3 = \zeta^2 A(\vec{\zeta}) \\ A(\vec{\zeta})^4 = \zeta^2 A(\vec{\zeta})^2 \\ \Lambda(\vec{\zeta}) = e^{A(\vec{\zeta})} = I + \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2 \\ \Lambda(-\vec{\zeta}) = I - \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2 \\ \Lambda(\vec{\zeta}) \Lambda(-\vec{\zeta}) = \left( I + \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2 \right) \left( I - \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2 \right) \\ \quad \quad \quad = I + \frac{ 2 \cosh \sqrt{ \zeta^2 } - 2 - \sinh^2 \sqrt{ \zeta^2 } + \cosh^2 \sqrt{ \zeta^2 } - 2 \cosh \sqrt{ \zeta^2 } + 1}{ \zeta^2} A(\vec{\zeta})^2 \\ \quad \quad \quad = I \) \(\vec{\theta} \to B(\vec{\theta}) = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -\theta_z & \theta_y \\ 0 & \theta_z & 0 & - \theta_x \\ 0 & -\theta_y & \theta_x & 0 \end{pmatrix} \\ B(\vec{\theta})^2 = \begin{pmatrix}0 & 0 & 0 & 0 \\ 0 & \theta_x^2 - \theta^2 & \theta_x \theta_y & \theta_x \theta_z \\ 0 & \theta_x \theta_y & \theta_y^2 - \theta^2 & \theta_y \theta_z \\ 0 & \theta_x \theta_z & \theta_y \theta_z & \theta_z^2 - \theta^2\end{pmatrix} \\ B(\vec{\theta})^3 = -\theta^2 B(\vec{\theta}) \\ B(\vec{\theta})^4 = -\theta^2 B(\vec{\theta})^2 \\ R(\vec{\theta}) = e^{B(\vec{\theta})} = I + \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \\ \quad \quad \quad = I + \frac{ \sinh \sqrt{ -\theta^2 } }{ \sqrt{ -\theta^2 } } B(\vec{\theta}) + \frac{ \cosh \sqrt{ -\theta^2 } - 1}{ -\theta^2} B(\vec{\theta})^2 \\ R(-\vec{\theta}) = I - \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \\ R(\vec{\theta}) R(-\vec{\theta}) = \left( I + \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \right) \left( I - \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \right) \\ \quad \quad \quad = I + \frac{ 2 - 2 \cos \sqrt{ \theta^2 } - \sin^2 \sqrt{ \theta^2 } - 1 + 2 \cos \sqrt{ \theta^2 } - \cos^2 \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \\ \quad \quad \quad = I \)
Define: \( \left[ \vec{u}_{\times} \right] \equiv \begin{pmatrix} 0 & - u_z & u_y \\ u_z & 0 & - u_x \\ - u_y & u_x & 0 \end{pmatrix} \\ \vec{u} \cdot \vec{v} \equiv \vec{u}^{\textrm{T}} \, \vec{v} = u_x v_x + u_y v_y + u_z v_z \\ u^2 \equiv \vec{u} \cdot \vec{u} \\ \vec{u} \otimes \vec{v} \equiv \vec{u} \, \vec{v}^{\textrm{T}} = \begin{pmatrix} u_x v_x & u_x v_y & u_x v_z \\ u_y v_x & u_y v_y & u_y v_z \\ u_z v_x & u_z v_y & u_z v_z \end{pmatrix} \\ A(\vec{u}) \equiv \begin{pmatrix} 0 & \vec{u}^{\textrm{T}} \\ \vec{u} & 0 \end{pmatrix} = \begin{pmatrix} 0 & u_x & u_y & u_z \\ u_x & 0 & 0 & 0 \\ u_y & 0 & 0 & 0 \\ u_z & 0 & 0 & 0 \end{pmatrix} \\ B(\vec{u}) \equiv \begin{pmatrix} 0 & \vec{0}^{\textrm{T}} \\ \vec{0} & \left[ \vec{u}_{\times} \right] \end{pmatrix} = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -u_z & u_y \\ 0 & u_z & 0 & - u_x \\ 0 & -u_y & u_x & 0 \end{pmatrix} \\ M( \vec{\rho} , \vec{\theta}) \equiv A(\vec{\rho}) + B(\vec{\theta}) = \begin{pmatrix} 0 & \vec{\rho}^{\textrm{T}} \\ \vec{\rho} & \left[ \vec{\theta}_{\times} \right] \end{pmatrix} = \begin{pmatrix} 0 & \rho_x & \rho_y & \rho_z \\ \rho_x & 0 & -\theta_z & \theta_y \\ \rho_y & \theta_z & 0 & - \theta_x \\ \rho_z & -\theta_y & \theta_x & 0 \end{pmatrix} \) \( A(\vec{\rho})^2 = \begin{pmatrix} \rho^2 & \vec{0}^{\textrm{T}} \\ \vec{0} & \vec{\rho} \otimes \vec{\rho} \end{pmatrix} \\ A(\vec{\rho})^3 = \rho^2 A(\vec{\rho}) \\ e^{A(\vec{\rho})} = I_4 + \sum_{k=0}^{\infty} \frac{ 1 }{ (2k+1)! } A(\vec{\rho})^{2k+1} + \sum_{k=0}^{\infty} \frac{ 1 }{ (2k+2)! } A(\vec{\rho})^{2k+2} \\ \quad \quad \quad = I_4 + \sum_{k=0}^{\infty} \frac{ \rho^{2k} }{ (2k+1)! } A(\vec{\rho}) + \sum_{k=0}^{\infty} \frac{ \rho^{2k} }{ (2k+2)! } A(\vec{\rho})^{2} \\ \quad \quad \quad = I_4 + \frac{\sinh \rho}{\rho} A(\vec{\rho}) + \frac{\cosh \rho - 1}{\rho^2} A(\vec{\rho})^{2} \) \(B(\vec{\theta})^2 = \begin{pmatrix} 0 & \vec{0}^{\textrm{T}} \\ \vec{0} & \vec{\theta} \otimes \vec{\theta} - \theta^2 I_3 \end{pmatrix} \\ B(\vec{\theta})^3 = - \theta^2 B(\vec{\theta}) \\ e^{B(\vec{\theta})} = I_4 + \sum_{k=0}^{\infty} \frac{ 1 }{ (2k+1)! } B(\vec{\theta})^{2k+1} + \sum_{k=0}^{\infty} \frac{ 1 }{ (2k+2)! } B(\vec{\theta})^{2k+2} \\ \quad \quad \quad = I_4 + \sum_{k=0}^{\infty} \frac{ (-1)^k \theta^{2k} }{ (2k+1)! } B(\vec{\theta}) + \sum_{k=0}^{\infty} \frac{ (-1)^k \theta^{2k} }{ (2k+2)! } B(\vec{\theta})^{2} \\ \quad \quad \quad = I_4 + \frac{\sinh \sqrt{-\theta^2} }{ \sqrt{-\theta^2} } B(\vec{\theta}) + \frac{\cosh \sqrt{-\theta^2} - 1}{- \theta^2} B(\vec{\theta})^{2} \\ \quad \quad \quad = I_4 + \frac{\sin \theta}{\theta} B(\vec{\theta}) + \frac{1 - \cos \theta}{\theta^2} B(\vec{\theta})^{2} \) \( A(\vec{\rho}) A(\vec{\theta}) = \begin{pmatrix} \vec{\rho} \cdot \vec{\theta} & \vec{0} ^{\textrm{T}} \\ \vec{0} & \vec{\rho} \otimes \vec{\theta} \end{pmatrix} \\ A(\vec{\theta}) A(\vec{\rho}) = \left( A(\vec{\rho}) A(\vec{\theta}) \right)^{\textrm{T}} \\ B(\vec{\rho}) B(\vec{\theta}) = \begin{pmatrix} \vec{0} & \vec{0} ^{\textrm{T}} \\ \vec{0} & \vec{\theta} \otimes \vec{\rho} - ( \vec{\rho} \cdot \vec{\theta} ) I_3\end{pmatrix} \\ B(\vec{\theta}) B(\vec{\rho}) = \left( B(\vec{\rho}) B(\vec{\theta}) \right)^{\textrm{T}} \\ A(\vec{\rho}) B(\vec{\rho}) = B(\vec{\rho}) A(\vec{\rho}) = 0 \\ A(\vec{\rho}) B(\vec{\theta}) = \begin{pmatrix} 0 & -( \vec{\theta} \times \vec{\rho} ) ^{\textrm{T}} \\ \vec{0} & 0 \end{pmatrix} \\ A(\vec{\rho}) B(\vec{\theta}) + A(\vec{\theta}) B(\vec{\rho}) = A(\vec{\rho}) B(\vec{\theta}) + A(\vec{\rho})^{\textrm{T}} B(\vec{\theta}) ^{\textrm{T}} = 0 \\ \left( A(\vec{\rho}) + B(\vec{\theta}) \right)\left( A(\vec{\theta}) - B(\vec{\rho}) \right) \\ \quad \quad \quad = A(\vec{\rho})A(\vec{\theta}) + B(\vec{\theta})A(\vec{\theta}) - A(\vec{\rho})B(\vec{\rho}) - B(\vec{\theta})B(\vec{\rho}) \\ \quad \quad \quad = A(\vec{\rho})A(\vec{\theta}) - B(\vec{\theta})B(\vec{\rho}) \\ \quad \quad \quad = ( \vec{\rho} \cdot \vec{\theta} ) I_4 \) \( \textrm{det} \, M(\vec{\rho}, \vec{\theta}) = - ( \vec{\rho} \cdot \vec{\theta} )^2 \\ M(\vec{\rho}, \vec{\theta})^{-1} = ( \vec{\rho} \cdot \vec{\theta} )^{-1} M(\vec{\theta}, -\vec{\rho}) ; \quad \textrm{if} \; \vec{\rho} \cdot \vec{\theta} \neq 0 \\ M(\vec{\rho}, \vec{\theta})^2 = \begin{pmatrix} \rho^2 & -( \vec{\theta} \times \vec{\rho} ) ^{\textrm{T}} \\ \vec{\theta} \times \vec{\rho} & \vec{\rho} \otimes \vec{\rho} + \vec{\theta} \otimes \vec{\theta} - \theta^2 I_3 \end{pmatrix} \\ M(\vec{\rho}, \vec{\theta})^3 = ( \rho^2 - \theta^2 ) M + ( \vec{\rho} \cdot \vec{\theta} )^2 M^{-1} \\ M(\vec{\rho}, \vec{\theta})^4 = (\rho^2 - \theta^2 ) M^2 + ( \vec{\rho} \cdot \vec{\theta} )^2 I_4 \) So the general case is going to be complicated, but there are interesting special cases.
This seems like a nicer way to think about it, decomposing into factors based on the trace and the traceless part. \(M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ \textrm{exp}(M) = e^{\frac{a+d}{2}} \left( \cosh \sqrt{ \frac{(a-d)^2}{4} + bc} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \frac{ \sinh \sqrt{ \frac{(a-d)^2}{4} + bc} }{ \sqrt{ \frac{(a-d)^2}{4} + bc} } \begin{pmatrix} \frac{a-d}{2} & b \\ c & \frac{d-a}{2} \end{pmatrix} \right)\)
