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?
Best Answer
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.
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.
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 $\cos(2t) + i\sin(2t)$.