# Applications of the Matrix exponential

Discussion in 'Physics & Math' started by rpenner, Feb 27, 2016.

1. ### rpennerFully WiredStaff Member

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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
$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ \textrm{exp}(M) = \frac{e^{\frac{a+d}{2}}}{\sqrt{(a-d)^2+4bc}} \begin{pmatrix} \sqrt{(a-d)^2+4bc} \cosh \frac{\sqrt{(a-d)^2+4bc}}{2} + (a-d) \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} & 2 b \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} \\ 2 c \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} & \sqrt{(a-d)^2+4bc} \cosh \frac{\sqrt{(a-d)^2+4bc}}{2} + (d-a) \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} \end{pmatrix}$

3. ### Confused2Registered Senior Member

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334
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.

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5. ### Dr_ToadIt's green!Valued Senior Member

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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!

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7. ### arfa branecall me arfValued Senior Member

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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?

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8. ### rpennerFully WiredStaff Member

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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.

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9. ### absolute-spaceRegistered Member

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ah, now I know who you are, sorry about the earlier comments,

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10. ### arfa branecall me arfValued Senior Member

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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 $A + B = \begin{pmatrix} 0 & 1 - i \\ 1 + i & 0 \end{pmatrix}$.

Last edited: Feb 28, 2016
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11. ### rpennerFully WiredStaff Member

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$M=\begin{pmatrix} 0 & 1 - i \\ 1 + i & 0 \end{pmatrix} = \begin{pmatrix} - \frac{1 -i}{2} & \frac{1 -i}{2} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} - \sqrt{2} & 0 \\ 0 & \sqrt{2} \end{pmatrix} \begin{pmatrix} - \frac{1 +i}{2} & \frac{1}{\sqrt{2}} \\ \frac{1 +i}{2} & \frac{1}{\sqrt{2}} \end{pmatrix}$

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:
$\textrm{exp}(M) = \begin{pmatrix} - \frac{1 -i}{2} & \frac{1 -i}{2} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} e^{- \sqrt{2}} & 0 \\ 0 & e^{\sqrt{2}} \end{pmatrix} \begin{pmatrix} - \frac{1 +i}{2} & \frac{1}{\sqrt{2}} \\ \frac{1 +i}{2} & \frac{1}{\sqrt{2}} \end{pmatrix} = \begin{pmatrix} \cosh \sqrt{2} & \frac{1 - i}{\sqrt{2}} \sinh \sqrt{2} \\ \frac{1 + i}{\sqrt{2}} \sinh \sqrt{2} & \cosh \sqrt{2} \end{pmatrix}$

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12. ### arfa branecall me arfValued Senior Member

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What about the expansion of $e^Ae^Be^{-A}e^{-B}$? Do A and B need to be Hermitian?

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13. ### rpennerFully WiredStaff Member

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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.

Last edited: Feb 28, 2016
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14. ### arfa branecall me arfValued Senior Member

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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: $\sigma_x \sigma_y = i \sigma_z$.

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15. ### rpennerFully WiredStaff Member

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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 $\sigma_x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_y = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} , \quad \sigma_z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$,

$e^{i \theta \sigma_x} = (\cos \theta) I + i (\sin \theta) \sigma_x \\ e^{i \theta \sigma_y} = (\cos \theta) I + i (\sin \theta) \sigma_y \\ e^{i \theta \sigma_z} = (\cos \theta) I + i (\sin \theta) \sigma_z \\ e^{-i\theta \sigma_y}e^{-i\theta \sigma_x}e^{i\theta \sigma_y}e^{i\theta \sigma_x} = ( \cos 2\theta + \frac{1}{2} \sin^2 2 \theta ) I + \frac{i}{2} (1 - \cos 2 \theta) (\sin 2 \theta) ( \sigma_x - \sigma_y) + \frac{i}{2} (\sin^2 2\theta) \sigma_z$

$\lim \limits_{\theta \to 0} e^{-i\theta \sigma_y}e^{-i\theta \sigma_x}e^{i\theta \sigma_y}e^{i\theta \sigma_x} \\ \quad = \left(1-2\theta^4+\frac{4}{3} \theta^6 \right) I + i \left( 2 \theta^3 - 2 \theta^5 + \frac{4}{5} \theta^7 \right) ( \sigma_x - \sigma_y) + i \left ( 2 \theta^2- \frac{8}{3} \theta^4 + \frac{64}{45} \theta^6 \right) \sigma_z$

$\begin{array}{c|ccc} \quad & \times \sigma_x & \times \sigma_y & \times \sigma_z \\ \hline \\ \sigma_x & I & i \sigma_z & - i \sigma_y \\ \sigma_y & -i \sigma_z & I & i \sigma_x \\ \sigma_z & i \sigma_y & -i \sigma_x & I \end{array}$

//
$e^{i \theta_x \sigma_x + i \theta_y \sigma_y + i \theta_z \sigma_z} = \left( \cos \sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2} \right) I + i \frac{\theta_x}{\sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2}} \left( \sin \sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2} \right) \sigma_z + i \frac{\theta_y}{\sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2}} \left( \sin \sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2} \right) \sigma_y + i \frac{\theta_z}{\sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2}} \left( \sin \sqrt{\theta_x^2 + \theta_y^2 + \theta_z^2} \right) \sigma_z$

Last edited: Mar 26, 2016
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16. ### rpennerFully WiredStaff Member

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Some special cases I worked out long-hand for another thread:

$\vec{\zeta} \to A(\vec{\zeta}) = \begin{pmatrix}0 & \zeta_x & \zeta_y & \zeta_z \\ \zeta_x & 0 & 0 & 0 \\ \zeta_y & 0 & 0 & 0 \\ \zeta_z & 0 & 0 & 0 \end{pmatrix} \\ A(\vec{\zeta})^2 = \begin{pmatrix}\zeta^2 & 0 & 0 & 0 \\ 0 & \zeta_x^2 & \zeta_x \zeta_y & \zeta_x \zeta_z \\ 0 & \zeta_x \zeta_y & \zeta_y^2 & \zeta_y \zeta_z \\ 0 & \zeta_x \zeta_z & \zeta_y \zeta_z & \zeta_z^2 \end{pmatrix} \\ A(\vec{\zeta})^3 = \zeta^2 A(\vec{\zeta}) \\ A(\vec{\zeta})^4 = \zeta^2 A(\vec{\zeta})^2 \\ \Lambda(\vec{\zeta}) = e^{A(\vec{\zeta})} = I + \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2 \\ \Lambda(-\vec{\zeta}) = I - \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2 \\ \Lambda(\vec{\zeta}) \Lambda(-\vec{\zeta}) = \left( I + \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2 \right) \left( I - \frac{ \sinh \sqrt{ \zeta^2 } }{ \sqrt{ \zeta^2 } } A(\vec{\zeta}) + \frac{ \cosh \sqrt{ \zeta^2 } - 1 }{ \zeta^2} A(\vec{\zeta})^2 \right) \\ \quad \quad \quad = I + \frac{ 2 \cosh \sqrt{ \zeta^2 } - 2 - \sinh^2 \sqrt{ \zeta^2 } + \cosh^2 \sqrt{ \zeta^2 } - 2 \cosh \sqrt{ \zeta^2 } + 1}{ \zeta^2} A(\vec{\zeta})^2 \\ \quad \quad \quad = I$

$\vec{\theta} \to B(\vec{\theta}) = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & -\theta_z & \theta_y \\ 0 & \theta_z & 0 & - \theta_x \\ 0 & -\theta_y & \theta_x & 0 \end{pmatrix} \\ B(\vec{\theta})^2 = \begin{pmatrix}0 & 0 & 0 & 0 \\ 0 & \theta_x^2 - \theta^2 & \theta_x \theta_y & \theta_x \theta_z \\ 0 & \theta_x \theta_y & \theta_y^2 - \theta^2 & \theta_y \theta_z \\ 0 & \theta_x \theta_z & \theta_y \theta_z & \theta_z^2 - \theta^2\end{pmatrix} \\ B(\vec{\theta})^3 = -\theta^2 B(\vec{\theta}) \\ B(\vec{\theta})^4 = -\theta^2 B(\vec{\theta})^2 \\ R(\vec{\theta}) = e^{B(\vec{\theta})} = I + \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \\ \quad \quad \quad = I + \frac{ \sinh \sqrt{ -\theta^2 } }{ \sqrt{ -\theta^2 } } B(\vec{\theta}) + \frac{ \cosh \sqrt{ -\theta^2 } - 1}{ -\theta^2} B(\vec{\theta})^2 \\ R(-\vec{\theta}) = I - \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \\ R(\vec{\theta}) R(-\vec{\theta}) = \left( I + \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \right) \left( I - \frac{ \sin \sqrt{ \theta^2 } }{ \sqrt{ \theta^2 } } B(\vec{\theta}) + \frac{ 1 - \cos \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \right) \\ \quad \quad \quad = I + \frac{ 2 - 2 \cos \sqrt{ \theta^2 } - \sin^2 \sqrt{ \theta^2 } - 1 + 2 \cos \sqrt{ \theta^2 } - \cos^2 \sqrt{ \theta^2 } }{ \theta^2} B(\vec{\theta})^2 \\ \quad \quad \quad = I$

17. ### rpennerFully WiredStaff Member

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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.

Last edited: May 5, 2016
18. ### rpennerFully WiredStaff Member

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This seems like a nicer way to think about it, decomposing into factors based on the trace and the traceless part.
$M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ \textrm{exp}(M) = e^{\frac{a+d}{2}} \left( \cosh \sqrt{ \frac{(a-d)^2}{4} + bc} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \frac{ \sinh \sqrt{ \frac{(a-d)^2}{4} + bc} }{ \sqrt{ \frac{(a-d)^2}{4} + bc} } \begin{pmatrix} \frac{a-d}{2} & b \\ c & \frac{d-a}{2} \end{pmatrix} \right)$

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