I have a $(3 \times 3)$ matrix $$ Y = \begin{pmatrix} 0 & – e^{-i \theta} & 0 \\ e^{i \theta} & 0 & – e^{-i \theta} \\ 0 & e^{i \theta} & 0 \end{pmatrix} $$ for which I would like to calculate the matrix exponential $\exp(t Y) = I + t Y + \frac{t^2 Y^2}{2!} + \ldots $ If I let $z : = e^{i \theta}$, I have $$ Y^2 = \begin{pmatrix} – |z|^2 & 0 & |z|^2 \\ 0 & -2 |z|^2 & 0 \\ |z|^2 & 0 & – |z|^2 \end{pmatrix} \\ Y^3 = \begin{pmatrix} 0 & 2 \overline{z} |z|^2 & 0 \\ |z|^2 (-z – \overline{z}) & 0 & |z|^2 (z + \overline{z}) \\ 0 & -2z |z|^2 & 0 \end{pmatrix} $$ and $$ Y^4 = \begin{pmatrix} – \overline{z} |z|^2 (-z – \overline{z}) & 0 & – \overline{z} |z|^2 (z + \overline{z}) \\ 0 & 4 |z|^4 & 0 \\ z |z|^2 (-z- \overline{z}) & 0 & z |z|^2 (z+ \overline{z}) \end{pmatrix}. $$ Setting $|z| = 1$ and calculating the matrix exponential above up the fifth power $Y^5$, I got $$ \begin{pmatrix} 1 – \frac{t^2}{2!} + \frac{t^4}{4!} \overline{z} (z + \overline{z}) + \ldots & – t \overline{z} + \frac{t^3}{3!} (2 \overline{z}) – \frac{t^5}{5!} 4 \overline{z} + \ldots & \frac{t^2}{2!} – \frac{t^4}{4!} \overline{z} (z + \overline{z}) + \ldots \\ tz – \frac{t^3}{3!} (z + \overline{z}) + \frac{t^5}{5!} 2 (z + \overline{z}) + \ldots & 1 – \frac{2 t^2}{2!} + \frac{t^4}{4!} 4 + \ldots & – t \overline{z} + \frac{t^3}{3!} (z + \overline{z}) – \frac{t^5}{5!} 2 ( z+ \overline{z}) + \ldots \\ \frac{t^2}{2!} – \frac{t^4}{4!} z (z + \overline{z}) + \ldots & tz – \frac{t^3}{3!} 2 z + \frac{t^5}{5!} 4 z + \ldots & 1 – \frac{t^2}{2!} + \frac{t^4}{4!} z (z + \overline{z}) + \ldots
\end{pmatrix} $$
I think I must be able to rewrite this with the help of $\sin(x) = x – \frac{x^3}{3!} + \frac{x^5}{5!} – \ldots$ and $\cos(x) = 1 – \frac{x^2}{2!} + \frac{x^4}{4!} – \ldots$.
For example, if I look at the $a_{22}$ term above, I see that it is almost $\cos(t)$, except the numerical factors that don't work out. Also, the $a_{11}$ term is almost $\cos(t)$, except there appears a term $\overline{z} (z+ z)$ from the fourth power onwards, and the same happens for the $a_{33}$ term with $z$ and $\overline{z}$ switched. The $a_{32}$ term seems to be $z \sin(t)$, but again the numerical coefficients don't work out.
Question: Does anyone recognize the pattern in these entries (i.e. the series) and is able to calculate the matrix exponential $e^{tY}$ in closed form?
Also, what would be matrix exponential $\exp(tZ)$ of the 'generalization' $$Z = \begin{pmatrix} 0 & – \overline{z} & – \overline{z} \\ z & 0 & – \overline{z} \\ z & z & 0 \end{pmatrix} $$ with $z = e^{i \theta}$ again?
Best Answer
Setting $z = e^{i \theta}$ is a good idea. It becomes a bit clearer if $(- e^{-i \theta})$ is replaced by $-1/z$ instead of $-\overline z$ (and it makes the result correct even for complex $\theta$).
So we have $$ Y = \begin{pmatrix} 0 & -1/z & 0 \\ z & 0 & -1/z \\ 0 & z & 0 \end{pmatrix} $$ and the first powers are $$ Y^2 = \begin{pmatrix} -1 & 0 & 1/z^2 \\ 0 & -2 & 0 \\ z^2 & 0 & -1 \end{pmatrix}\, , \, Y^3 = \begin{pmatrix} 0 & 2/z & 0 \\ -2z & 0 & 2/z \\ 0 & -2z & 0 \end{pmatrix}\,. \\ $$ One can see that $\boxed{Y^3 = -2Y}$, which allows to compute all powers $Y^n$ in terms of $Y$ or $Y^2$: $$ Y^{2k+1} = (-2)^{k} Y \\ Y^{2k+2} = (-2)^{k} Y^2 $$ for $k \ge 1$. Therefore $$ \begin{align} \exp(tY) &= I + \left(t-\frac{2t^3}{3!} + \frac{2^2t^5}{5!} - \frac{2^3t^7}{7!} + \ldots\right)Y \\ &\quad + \left(\frac{t^2}{2!} - \frac{2t^4}{4!} + \frac{2^2t^6}{6!} - \frac{2^3t^8}{8!} + \ldots \right)Y^2 \\ &= I + \frac{\sin(\sqrt 2 t)}{\sqrt 2}Y + \frac 12 \left(1- \cos(\sqrt 2 t)\right)Y^2 \, . \end{align} $$
The general case is described in Computing the Matrix Exponential The Cayley-Hamilton Method: If $A$ is an $n$-dimensional square matrix and $\lambda_1, \ldots, \lambda_n$ the zeros of the characteristic equation $\det(\lambda I - A) = 0$, then $$ \exp(tA) = \sum_{k_0}^{n-1} \alpha_k A^k $$ where $\alpha_0, \ldots, \alpha_{n-1}$ are the solutions of the linear equation system $$ e^{\lambda_i t} = \sum_{k_0}^{n-1} \alpha_k \lambda_i^k \, , \, 1 \le i \le n \, . $$
In our case $\det(\lambda I - Y) = \lambda^3 + 2 = 0$ has the zeros $\lambda_1 = 0$, $\lambda_2 = i\sqrt 2$, $\lambda_3 = -i \sqrt 2$. The linear equation system is $$ \begin{align} 1 &= \alpha_0 \\ e^{i\sqrt 2 t} &= \alpha_0 + i \sqrt 2 \alpha_1 - 2 \alpha_2 \\ e^{-i\sqrt 2 t} &= \alpha_0 - i \sqrt 2 \alpha_1 - 2 \alpha_2 \end{align} \, . $$ The solution is $$ \alpha_0 = 1, \, \alpha_1 = \frac{\sin(\sqrt 2 t)}{\sqrt 2}, \, \alpha_2 = \frac 12 \left(1- \cos(\sqrt 2 t)\right) $$ confirming the result for $\exp(tY)$ that we obtained above.