0
$\begingroup$

Do there exist matrices M and P for this equation? Or perhaps M and P dont need to be matrices? I saw this and this question after googling which made me wonder about whether the exponential form of complex numbers would still work.

$ z = a\begin{pmatrix} 1&0\\ 0&1 \end{pmatrix}+b\begin{pmatrix} 0&1\\ -1&0 \end{pmatrix} = \begin{pmatrix} a&b\\ -b&a \end{pmatrix} = Me^P $

$ $

$ $

Addition:

I just found this article which I dont really understand, but where they solve a case similar to this. The thing I dont understand in this article is how you exponentiate with a matrix and what that alpha is doing there. Also I dont understand that in order to make this work they have to make the matrix $\Phi$ a complex matrix, looks like it defeats the purpose of having a matrix representation of complex numbers. Not sure whether to append this to my question or add as answer, because I still dont understand it.

They say, $ z = \begin{pmatrix} a&jb\\ j\alpha^2b&a \end{pmatrix} = aI+b\Phi $

Where, $ \Phi = \begin{pmatrix} 0&j\\ j\alpha^2&0 \end{pmatrix} $

For which they say, $ e^{\phi\Phi}=cos(\alpha\phi)I+\frac{1}{\alpha}sin(\alpha\phi)\Phi $

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Show that $z=\rho e^{i\theta}$ with $i\theta=\begin{pmatrix}0&\theta\\-\theta&0\end{pmatrix}$. Calculate $e^{i\theta}$ and deduce $\theta,\rho$.

$\endgroup$
7
  • $\begingroup$ Thanks! Looks like $\rho$ and $\theta$ should be scalars then, right? Could you explain how you calculate $e^{\begin{pmatrix}0&\theta\\-\theta&0\end{pmatrix}}$? $\endgroup$
    – Leo
    Commented Apr 16, 2014 at 13:47
  • 1
    $\begingroup$ Use the definition: $e^A=I+A+A^2/2+\cdots$ ; you'll see trigonometric functions appear. $\endgroup$
    – user91684
    Commented Apr 16, 2014 at 13:50
  • $\begingroup$ Ah yes, it comes out nicely! Exactly nothing changes in how I calculate the magnitude and the phase as with normal notation: I ended up doing exactly the same as before. Is there a linear algebra way to calculate the "magnitude" and "phase" of $\begin{pmatrix} a&b\\ -b&a \end{pmatrix}$ ? $\endgroup$
    – Leo
    Commented Apr 16, 2014 at 14:20
  • 1
    $\begingroup$ Let $Z=\begin{pmatrix}a&b\\-b&a\end{pmatrix}$. Then $\rho=\sqrt{\det(Z)}$ and $Z=\rho O$ where $O\in O^+(2)$. Bourbaki defines an angle as an element of $O^+(2)$, identifying $\theta+2\pi\mathbb{Z}$ and $e^{i\theta}$. The effective calculation of $\theta$ is an analytic calculation. $\endgroup$
    – user91684
    Commented Apr 16, 2014 at 21:09
  • 1
    $\begingroup$ $O⁺(n)$ is the group of rotations in dimension $n$ (orthogonal matrices with determinant $1$). $\endgroup$
    – user91684
    Commented Apr 17, 2014 at 12:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .