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This is part of a much larger ODE problem and I want to know how to compute $\exp(Bt)$. Now one method is the brute force, that is calculating eigenvectors of $B$ corresponding to the two imaginary eigenvalues. Then construct the matrix $P$ consisting of the eigenvectors as columns. We can then calculate the inverse $P^{-1}$ and conclude that

$$\exp(Bt)=P\exp(Dt)P^{-1}$$

where $D$ is a diagonal matrix where the diagonals are the eigenvalues of $B$. doing this I get the correct asnwer, however this is tedious and in the notes they recommend the following:

Multiplication and addition of matrices of the form $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$ obey the same rules as multiplication and addition of complex numbers $z=a+bi$. There fore the matrix $\begin{bmatrix}0&-b\\b&0\end{bmatrix}$ corresponds through this correspondence to purely imaginary numbers, and the relation $\exp(ib)=\cos{b}+i\sin{b}$ can be applied leading to the answer

$$\exp(Bt)=\begin{bmatrix}\cos{2t}&-\sin{2t}\\\sin{2t}&\cos{2t}\end{bmatrix}.$$

Can someone show me what is happening here?

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    $\begingroup$ This would make sense if $B=\left[\begin{smallmatrix}0&-2\\2&0\end{smallmatrix}\right]$. Are you sure that you wrote it right? $\endgroup$ – José Carlos Santos Jun 1 at 20:39
  • $\begingroup$ @JoséCarlosSantos - Corrected! Can you show me how it makes sense? $\endgroup$ – Parseval Jun 1 at 20:45
  • $\begingroup$ No need for that, since Arturo Magidin has already provided a fine answer. $\endgroup$ – José Carlos Santos Jun 1 at 20:49
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Set

$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag 1$

then it is easily seen that

$J^2 = -I; \tag 2$

continuing in this manner we find

$J^3 = -J, \tag 3$

$J^4 = -J^2 = I, \tag 4$

$J^5 = J, \tag 5$

and in general for $n \in \Bbb N$,

$J^{4n + k} = J^{4n}J^k = (J^n)^4J^k = J^k, \tag 6$

where

$0 \le k < 4; \tag 7$

we only need consider $k$ in this limited range since every $m \in \Bbb N$ may be uniquely written as

$m = 4n + k, \; n \in \Bbb N, 0 \le k < 4 \tag 8$

by the classical Euclidean division algorithm. The reader will undoubtedly recognize the pattern in the powers of $J$ as identical to that of the powers of $i \in \Bbb C$:

$i^2 = -1, \; i^3 = -i, \; i^4 = 1, \tag 9$

and of course

$i^{4n + k} = (i^4)^n i^k = i^k; \tag{10}$

furthermore, the matrix $J$ is of unit norm,

$\Vert J \Vert = 1, \tag{11}$

which is readily evident since via (1)

$\left \Vert J\begin{pmatrix} a \\ b \end{pmatrix} \right \Vert^2 = \left \Vert \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{pmatrix} a \\ b \end{pmatrix} \right \Vert^2 = \left \Vert \begin{pmatrix} -b \\ a \end{pmatrix} \right \Vert^2 = b^2 + a^2 = \left \Vert \begin{pmatrix} a \\ b \end{pmatrix} \right \Vert^2,\tag{12}$

whence

$\left \Vert J\begin{pmatrix} a \\ b \end{pmatrix} \right \Vert = \left \Vert \begin{pmatrix} a \\ b \end{pmatrix} \right \Vert, \tag{13}$

which implies (11).

Now for any

$\omega \in \Bbb R \tag{14}$

we have the power series representation of

$e^{i\omega t} = \displaystyle \sum_0^\infty \dfrac{(i\omega t)^n}{n!}; \tag{15}$

using (9)-(10), this series may be re-arranged as follows:

$\displaystyle \sum_0^\infty \dfrac{(i\omega t)^n}{n!} = \sum_0^\infty \dfrac{(\omega t)^n i^n}{n!}$ $= \displaystyle \sum_0^\infty \dfrac{(\omega t)^{2n}i^{2n}}{(2n)!} + \sum_0^\infty \dfrac{(\omega t)^{2n + 1}i^{2n + 1}}{(2n + 1)!} = \sum_0^\infty \dfrac{(\omega t)^{2n}i^{2n}}{(2n)!} + i\sum_0^\infty \dfrac{(\omega t)^{2n + 1}i^{2n}}{(2n + 1)!}$ $= \displaystyle \sum_0^\infty \dfrac{(-1)^n(\omega t)^{2n}8i}{(2n)!} + i\sum_0^\infty \dfrac{(-1)^n (\omega t)^{2n + 1}}{(2n + 1)!}. \tag{16}$

Those familiar with the Taylor series of basic trigonometric functions will recognize the expression on the extreme right of (16) as $\cos(\omega t) + i\sin(\omega t)$; that is, we have shown that

$e^{¡ \omega t} = \cos(\omega t) + i\sin(\omega t). \tag{17}$

A word about convergence: all of the above series converge absolutely as may be seen via the ratio test; we illustrate this for the last series on the extreme right of (16), that is, for

$\sin(\omega t) = \displaystyle \sum_0^\infty \dfrac{(-1)^n (\omega t)^{2n + 1}}{(2n + 1)!}; \tag{18}$

the ratio of the absolute values of succeeding terms is then

$\rho_n = \dfrac{\vert \omega t \vert^{2n + 3}}{(2n + 3)!} / \dfrac{\vert \omega t \vert^{2n + 1}}{(2n + 1)!} = \dfrac{\vert \omega t \vert^2}{(2n + 2)(2n + 3)} \to 0 \; \text{as} \; n \to \infty, \tag{19}$

independently of $\omega$ and $t$; it is the absolute convergence of these series which ensures the legitimacy of the re-arrangements and re-groupings performed in (16).

We next observe that, by virtue of (2)-(8), every step of the calculation (16) carries through with $i$ replaced by $J$; also, (11) allows us to conclude that the resulting series are also absolutely convergent, and therefore we may write

$e^{\omega t J} = \cos (\omega t) I + \sin (\omega t) J. \tag{20}$

Finally, taking

$\omega = 2, \tag{21}$

we see that

$e^{Bt} = e^{2 t J} = \cos(2t) I + \sin(\omega t) J = \begin{bmatrix} \cos (2t) & -\sin(2t) \\ \sin(2t) & \cos (2t) \end{bmatrix}, \tag{22}$

as was to be shown.

And that is what is happening here!

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  1. Note that $$\left(\begin{array}{cr} a & -b\\ b & a\end{array}\right) + \left(\begin{array}{cr}c & -d\\ d& c\end{array}\right) = \left(\begin{array}{cr} a+c & -(b+d)\\ (b+d) & a+c \end{array}\right)$$ and $$\left(\begin{array}{cr} a & -b\\ b & a\end{array}\right) \left(\begin{array}{cr}c & -d\\ d& c\end{array}\right) = \left(\begin{array}{cr} ac-bd & -(ad+bc)\\ ad+bc & ac-bd \end{array}\right).$$ So if you identify the complex number $a+bi$ with the matrix $\left(\begin{array}{cr} a&-b\\ b&a\end{array}\right)$ you have that the sum of the images is the image of the sum, and the product of the images is the image of the product. That is, these matrices give you a way to “code” complex numbers as $2\times 2$ matrices.

  2. Because both matrix exponential and complex exponential are defined in terms of the same power series, if the matrix $M$ is of the above form and corresponds to the complex number $z$, then you get that $$\exp(z) = \sum_{i=0}^{\infty}\frac{z^n}{n!}$$ must correspond to the matrix $$\sum_{i=0}^{\infty} \frac{M^n}{n!} = \exp{M}.$$ This is a bit “fluffy”, since you are really talking about convergence, and it is only the partial sums that correspond to one another, but in point of fact it turns out that it does work as one hopes.

  3. The matrix you have thus corresponds to the complex number $2i$. The matrix $Bt$ corresponds to the complex number $2ti$. Hence, $\exp(Bt)$ must correspond to the complex number $\exp(2ti) = \cos(2t) + i\sin(2t)$, and hence you get $$ \exp(Bt) = \left(\begin{array}{cr} \cos(2t) & -\sin(2t)\\ \sin(2t) & \cos(2t)\end{array}\right),$$ because that’s the matrix that corresponds to the complex number $\cos(2t) + i\sin(2t)$.

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