Matrix exponential of a traceless $2\times2$ complex matrix

Question

How to calculate $\exp\left(t\begin{bmatrix}0 & z\\z^* & 0\end{bmatrix}\right)$, or $\exp\left(\begin{bmatrix}0 & v\\w & 0\end{bmatrix}\right)$ (where: $v, w \in \mathbb{C}$) in general?

Answer 1

$e^t=1+t+\frac{t^2}{2 !}+\ldots,$ so for any matrix $A$, $e^A=I+A+\frac{1}{2!}A^2+\ldots$.

Answer 2

By definition $e^A=\sum_{j=0}^\infty \frac{A^j}{j!}$ where $A^0=I\equiv$the identity matrix. Now you have $A={\begin{bmatrix}0 & v\\w & 0\end{bmatrix}}$. Note that $A^2=vwI$, so by induction you can prove $A^{2j}=v^jw^jI$ and $A^{2j+1}=v^jw^jA$ for all $j\ge 0$. It follows that \begin{align} e^A&=\sum_{j=0}^\infty \frac{A^j}{j!}\\ &=\sum_{j=0}^\infty \frac{A^{2j}}{(2j)!}+\sum_{j=0}^\infty \frac{A^{2j+1}}{(2j+1)!}\\ &=\sum_{j=0}^\infty \frac{v^jw^jI}{(2j)!}+\sum_{j=0}^\infty \frac{v^jw^jA}{(2j+1)!}\\ &=\sum_{j=0}^\infty \frac{v^jw^j}{(2j)!}I+\sum_{j=0}^\infty \frac{v^jw^j}{(2j+1)!}A\\ &={\begin{bmatrix}\sum_{j=0}^\infty \frac{v^jw^j}{(2j)!} & \sum_{j=0}^\infty \frac{v^{j+1}w^j}{(2j+1)!}\\\sum_{j=0}^\infty \frac{v^jw^{j+1}}{(2j+1)!} & \sum_{j=0}^\infty \frac{v^jw^j}{(2j)!}\end{bmatrix}}\\ \end{align}

From this point, I am sure you can evaluate each element of the matrix in the last line, can't you?

Answer 3

Let $A=\pmatrix{0&v\\ w&0}$ and $x$ be any complex number such that $vw=-x^2$. By Cayley-Hamilton theorem, $A^2 = \operatorname{trace}(A)A-\det(A)I = vwI = -x^2I$. Therefore, when $vw\ne0$ (so that $x\ne0$), \begin{align*} e^A &= I+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\ldots\\ &= \left(I+\frac{A^2}{2!}+\frac{A^4}{4!}+\ldots\right) +\left(A+\frac{A^3}{3!}+\frac{A^5}{5!}+\ldots\right)\\ &= \left(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\ldots\right)I +\left(1-\frac{x^2}{3!}+\frac{x^4}{5!}-\ldots\right)A\\ &= \cos(x)I+\frac{\sin(x)}{x}A.\tag{1} \end{align*} Since the exponential function is continuous, when $vw=0$, we can take the limit of $(1)$ with $x\to0$ to obtain that $e^A= I+A$.

Answer 4

Of course, Naveen's answer is correct insofar as we can exponentiate any matrix $A$ by computing the power series $\sum_0^\infty (A^n/n!)$, direct application of this technique involves very many calculations; it is truly a lot work. But the special form of the matrices given in the question allow some extensive simplifications to me made, and a pretty, elegant solution may be had. To wit:

We will first calculate $e^W$, where

$W = \begin{bmatrix} 0 & v \\ w & 0 \end{bmatrix}. \tag{1}$

We have:

$W^2 = \begin{bmatrix} 0 & v \\ w & 0 \end{bmatrix} \begin{bmatrix} 0 & v \\ w & 0 \end{bmatrix} = \begin{bmatrix} vw & 0 \\ 0 & vw \end{bmatrix} = vwI, \tag{2}$

$W^3 = W^2W = vwIW = vwW, \tag{3}$

$W^4 = (W^2)^2 = (vwI)^2 = v^2w^2I = (vw)^2I, \tag{4}$

$W^5 = W^4W = v^2w^2IW = (vw)^2W; \tag{6}$

the general pattern emerges from (2)-(6); it is

$W^{2n} = v^nw^nI = (vw)^nI, \tag{7}$

$W^{2n + 1} = W^{2n}W = v^nw^nW = (vw)^nW; \tag{8}$

(7)-(8) may be easily validated by an extremely simple induction; indeed, if

$W^{2k} = (vw)^kI, \tag{9}$

then

$W^{2(k + 1)} = W^{2k + 2} = W^{2k}W^2 = ((vw)^kI)(vwI) = (vw)^{k + 1}I; \tag{10}$

and if

$W^{2k + 1} = v^kw^kW = (vw)^kW, \tag{11}$

then

$W^{2(k + 1) + 1} = W^{2k + 3} = W^{2k + 1}W^2 = ((vw)^kW)(vwI) = (vw)^{k + 1}W; \tag{12}$

thus we see that (7)-(8) are valid for all $n \ge 0$. Based on these relations, we compute $e^W$ as

$e^W = \sum_0^\infty \dfrac{W^n}{n!} = \sum_0^\infty \dfrac{W^{2n}}{(2n)!} + \sum_0^\infty \dfrac{W^{2n + 1}}{(2n + 1)!}, \tag{13}$

where we have broken the series into even and odd powers of $W$. We have, using (7), (8):

$\sum_0^\infty \dfrac{W^{2n}}{(2n)!} = \sum_0^\infty \dfrac{(vw)^nI}{(2n)!} = (\sum_0^ \infty \dfrac{(vw)^n}{(2n)!})I, \tag{14}$

and

$\sum_0^\infty \dfrac{W^{2n + 1}}{(2n + 1)!} = \sum_0^\infty \dfrac{(vw)^nW}{(2n + 1)} = (\sum_0^\infty \dfrac{(vw)^n}{(2n + 1)!})W; \tag{15}$

if we now choose $r \in \Bbb C$ such that $r^2 = vw$, and re-assemble (13) using (14) and (15) with $r^2$ in place of $vw$, obtaining

$e^W = (\sum_0^\infty \dfrac{r^{2n}}{(2n!)})I + (\sum_0^\infty \dfrac{r^{2n}}{(2n + 1)!})W. \tag{16}$

We now branch and treat the two cases $r = 0$, $r \ne 0$ separately. First, if $r \ne 0$, we can re-write (16) in the form

$e^W = (\sum_0^\infty \dfrac{r^{2n}}{(2n)!})I+ (\sum_0^\infty \dfrac{r^{2n + 1}}{(2n + 1)!})\dfrac{W}{r}. \tag{17}$

We now observe that the first sum in (17) consists precisely of the even degree terms of

$e^r = \sum_0^\infty \dfrac{r^n}{n!}, \tag{18}$

as such, it is in fact $\cosh r = (e^r + e^{-r})/2$, in which the odd degree terms all cancel out; likewise the second sum is $\sinh r = (e^r - e^{-r})/2$, the terms of even degree cancelling in this case. Thus

$e^W = (\cosh r) I + (\sinh r) \dfrac{W}{r}; \tag{19}$

in the event that $r = 0$, we see directly from (16) that

$e^W = I + W, \tag{20}$

which is consistent with the limit of (19) as $r \to 0$, since $\lim_{r \to 0} \sinh r/ r = 1$, as may readily seen from the definition $\sinh r = (e^r - e^{-r})/2$. Indeed, the function $\sigma(r) = \sinh r /r$ is a well-defined analytic function for all $r$, as may be seen by inspection of the power series for $\sinh r / r$:

$\sigma(r) = \dfrac{\sinh r}{r} = \sum_0^\infty \dfrac{r^{2n}}{(2n + 1)!}. \tag{21}$

It is evident that (19) is both a structurally illuminating and a computationally efficient representation of $e^W$, since it not only presents $e^W$ in a form which exposes different aspects of its expansion as a linear combination of powers of $W$, but also represents a method for evaluating $e^W$ which requires far less effort than direct matrix expansion of the power series $e^W$ as suggested by Naveen. Having made note of these things, we proceed.

With the formulas (19) and (20) at our disposal, setting

$W = t\begin{bmatrix} 0 & z \\ z^\ast & 0 \end{bmatrix} = \begin{bmatrix} 0 & tz \\ tz^\ast & 0 \end{bmatrix} = tZ, \tag{22}$

where

$Z = \begin{bmatrix} 0 & z \\ z^\ast & 0 \end{bmatrix}; \tag{23}$

then with $v = tz$ and $w = tz^\ast$, we have $r^2 = t^2zz^\ast = t^2 \vert z \vert^2$, so that we may take $r = t \vert z \vert$. Then (19) yields

$e^{tZ} = (\cosh t \vert z \vert) I + (\sinh t \vert z \vert) \dfrac{tZ}{t \vert z \vert} = (\cosh t \vert z \vert) I + (\sinh t \vert z \vert) \dfrac{Z}{\vert z \vert}. \tag{24}$

A few final remarks concerning the choice of $r = t \vert z \vert$: the reader will recall from the above that $r$ only need satisfy $r^2 = vw$ or $r^2 = \vert t^2 zz^\ast \vert$; clearly, taking $r = t \vert z \vert$ meets this requirement, as do the choices $r = -t \vert z \vert$, $r = \vert tz \vert$, $r = - \vert tz \vert$ etc. In fact, the slection of $r$ doesn't even have to be continuous with respect to $u, v$ or $t, \vert z \vert$, but $r^2$ does. I chose $r = t\vert z \vert$ because, out of all these choices, it makes the $t$-dependence most clear. At least to Yours Truly.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!