Applications of the Matrix exponential

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

  1. rpenner Fully Wired Staff 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
    \(M = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \\ \textrm{exp}(M) = \frac{e^{\frac{a+d}{2}}}{\sqrt{(a-d)^2+4bc}} \begin{pmatrix} \sqrt{(a-d)^2+4bc} \cosh \frac{\sqrt{(a-d)^2+4bc}}{2} + (a-d) \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} & 2 b \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} \\ 2 c \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} & \sqrt{(a-d)^2+4bc} \cosh \frac{\sqrt{(a-d)^2+4bc}}{2} + (d-a) \sinh \frac{\sqrt{(a-d)^2+4bc}}{2} \end{pmatrix} \)
     
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  3. Confused2 Registered Senior Member

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    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_Toad It'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!

    Please Register or Log in to view the hidden image!

     
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  7. arfa brane call me arf Valued 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. rpenner Fully Wired Staff 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-space Registered Member

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    ah, now I know who you are, sorry about the earlier comments,
     
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  10. arfa brane call me arf Valued 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. rpenner Fully Wired Staff 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 brane call me arf Valued 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. rpenner Fully Wired Staff 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 brane call me arf Valued 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. rpenner Fully Wired Staff 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. rpenner Fully Wired Staff 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. rpenner Fully Wired Staff 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. rpenner Fully Wired Staff 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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