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Trace of the Exponential of a Square Matrix

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  1. Dec 1, 2017 #1
    1. The problem statement, all variables and given/known data
    Find the trace of a ##4\times 4## matrix ##\mathbb U=exp(\mathbb A)##, where
    $$\mathbb A = \begin {pmatrix} 0&0&0&{\frac {\pi}{4}}\\ 0&0&{\frac {\pi}{4}}&0\\ 0&{\frac {\pi}{4}}&0&0\\ {\frac {\pi}{4}}&0&0&0 \end {pmatrix}$$

    2. Relevant equations
    $$e^{(\mathbb A)}=\mathbb P {e^{\mathbb D}} {\mathbb P}^{-1}$$

    3. The attempt at a solution
    Eigenvalues of ##\mathbb (A)## are ##{\lambda}_1=-\frac {\pi}{4}##, ##{\lambda}_2=-\frac {\pi}{4}##, ##{\lambda}_3=\frac {\pi}{4}## and ##{\lambda}_4=\frac {\pi}{4}##.
    The corresponding eigenvectors are
    $${\vec v}_1=\begin {pmatrix} 0\\ -1\\1\\0 \end {pmatrix}$$
    $${\vec v}_2=\begin {pmatrix} -1\\ 0\\0\\1 \end {pmatrix}$$
    $${\vec v}_3=\begin {pmatrix} 0\\ 1\\1\\0 \end {pmatrix}$$
    $${\vec v}_4=\begin {pmatrix} 1\\ 0\\0\\1 \end {pmatrix}$$

    So, $$\mathbb P = \begin {pmatrix} 0&-1&0&1\\ -1&0&1&0\\ 1&0&1&0\\ 0&1&0&1 \end {pmatrix}$$

    $$Co{\mathbb P} = \begin {pmatrix} 0&2&0&-2\\ 2&0&-2&0\\ -2&0&-2&0\\ 0&-2&0&-2 \end {pmatrix}$$

    $${(Co{\mathbb P})}^T = \begin {pmatrix} 0&2&-2&0\\ 2&0&0&-2\\ 0&-2&-2&0\\ -2&0&0&-2 \end {pmatrix}$$

    $$det{\mathbb P} = \begin {vmatrix} 0&-1&0&1\\ -1&0&1&0\\ 1&0&1&0\\ 0&1&0&1 \end {vmatrix}$$
    Which gives
    $$det{\mathbb P} = -4$$

    Now, $${\mathbb P}^{-1} = \frac {{(Co{\mathbb P})}^T}{det{\mathbb P}}$$
    So, $${\mathbb P}^{-1} = {\frac {1}{- 4}}{\begin {pmatrix} 0&2&-2&0\\ 2&0&0&-2\\ 0&-2&-2&0\\ -2&0&0&-2 \end {pmatrix}}$$

    After simplification
    $${\mathbb P}^{-1} = \begin {pmatrix} 0&{-\frac {1}{2}}&{\frac {1}{2}}&0\\ {-\frac {1}{2}}&0&0&{\frac {1}{2}}\\ 0&{\frac {1}{2}}&{\frac {1}{2}}&0\\ {\frac {1}{2}}&0&0&{\frac {1}{2}} \end {pmatrix}$$

    Then $${{\mathbb P}^{-1}}{\mathbb A}{\mathbb P} = {\begin {pmatrix} 0&{-\frac {1}{2}}&{\frac {1}{2}}&0\\ {-\frac {1}{2}}&0&0&{\frac {1}{2}}\\ 0&{\frac {1}{2}}&{\frac {1}{2}}&0\\ {\frac {1}{2}}&0&0&{\frac {1}{2}}\end {pmatrix}} {\begin {pmatrix} 0&0&0&{\frac {\pi}{4}}\\ 0&0&{\frac {\pi}{4}}&0\\ 0&{\frac {\pi}{4}}&0&0\\ {\frac {\pi}{4}}&0&0&0 \end {pmatrix}}{\begin {pmatrix} 0&-1&0&1\\ -1&0&1&0\\ 1&0&1&0\\ 0&1&0&1\end {pmatrix}}$$
    This simplifies, giving the diagonal matrix $${{\mathbb P}^{-1}}{\mathbb A}{\mathbb P} =\begin {pmatrix} -{\frac {\pi}{4}}&0&0&0\\ 0&-{\frac {\pi}{4}}&0&0\\ 0&0&{\frac {\pi}{4}}&0\\ 0&0&0&{\frac {\pi}{4}} \end {pmatrix}$$

    Now, I don't know if $$e^{{{\mathbb P}^{-1}}{\mathbb A}{\mathbb P}} =\begin {pmatrix} e^{-{\frac {\pi}{4}}}&0&0&0\\ 0&e^{-{\frac {\pi}{4}}}&0&0\\ 0&0&e^{{\frac {\pi}{4}}}&0\\ 0&0&0&e^{{\frac {\pi}{4}}} \end {pmatrix}$$

    If it is so, ##trace {[\mathbb P {e^{\mathbb D}} {\mathbb P}^{-1}]}## looks not ok.

    Can anyone suggest any simpler way to handle such a problem?
     
    Last edited: Dec 1, 2017
  2. jcsd
  3. Dec 1, 2017 #2

    DrClaude

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    Staff: Mentor

    This equation is not correct.
     
  4. Dec 1, 2017 #3
    Thanks @DrClaude, for pointing out.
    It should be $$e^{(\mathbb A)}=\mathbb P {e^{\mathbb D}}{\mathbb P}^{-1}$$
    I corrected it.
     
  5. Dec 1, 2017 #4

    DrClaude

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    Staff: Mentor

    You now need to calculate ##\mathbb P {e^{\mathbb D}}{\mathbb P}^{-1}##.
     
  6. Dec 1, 2017 #5
    $$\mathbb P {e^{\mathbb D}} {\mathbb P}^{-1}=\begin {pmatrix} \cosh {\frac {\pi}{4}}&0&0&\sinh {\frac {\pi}{4}}\\ 0&\cosh {\frac {\pi}{4}}&\sinh {\frac {\pi}{4}}&0\\ 0&\sinh {\frac {\pi}{4}}&\cosh {\frac {\pi}{4}}&0\\ \sinh {\frac {\pi}{4}}&0&0&\cosh {\frac {\pi}{4}} \end {pmatrix}$$

    So, I just take its ##trace##?
    That's all?
     
  7. Dec 1, 2017 #6

    Ray Vickson

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    Science Advisor
    Homework Helper

    Yes, but that is doing it the hard way. An often-easier approach is to note that if the eigenvalues of a ##4 \times 4## matrix ##B## are ##r_1, r_1, r_2, r_2## the general form of ##f(B)## for an analytic function ##f(\cdot)## is
    $$ f(B) = E_0 f(r_1) + E_1 f^{\prime}(r_1) + F_0 f(r_2) + F_1 f^{\prime}(r_2) \hspace{3ex}(1)$$
    for some fixed matrices ##E_0,E_1,F_0,F_1## that are the same for any function ##f##. Here, I take
    $$B = \pmatrix{0 & 0 & 0 & 1\\0 & 0 & 1 &0\\0 &1 & 0 & 0 \\ 1 & 0 & 0 & 0},$$
    so that we want to know ##f(B) = \exp( \pi/4 \: B)##.

    We can determine these matrices just by using equation (1) on the four functions ##f_0(x) = 1 \Rightarrow f_0(B) = I##, ##f_1(x) = x \Rightarrow f_1(B) = B##, ##f_2(x) = x^2 \Rightarrow f_2(B) = B^2## and ##f_3(x) = x^3 \Rightarrow f_3(B) = B^3##. In this case you have ##r_1 = 1## and ##r_2 = -1##, so that gives four equations:
    $$\begin{array}{ccl}
    I &=& E_0 + F_0\\
    B &=& E_0+E_1 -F_0 + F_1 \\
    B^2 &=& E_0 + 2 E_1 + F_0 - 2 F_1 \\
    B^3 &=& E_0 + 3 E_1 - F_0 + 3 F_1
    \end{array}
    $$
    You can solve these, and then get
    $$ e^{\pi/4 \: B} = E_0 e^{\pi/4} + (\pi/4) E_1 e^{\pi/4} + F_0 e^{-\pi/4} - (\pi/4) F_1 e^{-\pi/4}, $$
    so that
    $$\text{trace} \left(e^{\pi/4 \: B}\right) = \text{trace} (E_0) e^{\pi/4} + (\pi/4) \text{trace} (E_1) e^{\pi/4} + \text{trace}(F_0) e^{-\pi/4}
    -(\pi/4) \text{trace}(F_1) e^{-\pi/4}$$
     
    Last edited: Dec 1, 2017
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