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Given $a\in\Bbb F$ are there any natural $k\times k$ matrices in $\Bbb F^{k\times k}$ with $a^{0},a^1,a^2,\dots, a^{k-1}$ as eigenvalues where $\Bbb F$ is any char $0$ field?

The characteristic equation should have roots in G.P. as well. How should the char equation look like?

Can we conclude that all such matrices should be structurally similar and just look at circulants with these prescribed eigenvalues since other matrices are just conjugate by unitaries of diagonal matrices with specified numbers as roots? (Note circulants are rotations by FFT)

What is the matrix structure in general?

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    $\begingroup$ What kinds of matrices do you allow, $\text{Mat}_k\mathbb{C}$ or $\text{Mat}_k\mathbb{R}$? (After all, real matrices can have complex eigenvalues.) $\endgroup$ – Fly by Night Oct 11 '13 at 20:20
  • $\begingroup$ In $\mathbb{C}$, how about the matrices that look like $diag (a^0, a^1, \ldots a^{k-1})$? Are those 'natural'? $\endgroup$ – Calvin Lin Oct 11 '13 at 20:24
  • $\begingroup$ changed question $\endgroup$ – 1.. Oct 11 '13 at 20:24
  • $\begingroup$ @CalvinLin that is exactly the question. What are all the matrices? Are they just rotations of diagonlas? $\endgroup$ – 1.. Oct 11 '13 at 20:25
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    $\begingroup$ Well, you know that the eigenspaces are 1-dimensional, and you must have 1 vector in each of them, so that's a complete classification (at least in $\mathbb{C}$). $\endgroup$ – Calvin Lin Oct 11 '13 at 20:26
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If you have $k$-by-$k$ real matrices then the characteristic polynomial must be real. Assume that the eigenvalues are real and all have a common ratio. If they are all real then:

  • $\chi(s) \propto (s-v)^k$
  • $\chi(s) \propto (s-v)(s-rv)(s-r^2v)\ldots (s-r^kv)$

In the first case, the matrix is a multiple of the identity. In the second it is diagonalisable.

If there is a complex eigenvalue then its conjugate must also be an eigenvalue. If the common ratio is $r$ then there exists a positive integer $k$ for which $$z=r^kz^* \implies |z|=|r^kz^*| \implies |z|=|r|^k|z^*| \implies |r|^k=1 \implies |r|=1$$ It follows that the eigenvalues must all have the same modulus: $|r^kz|=|r|^k|z|=|z|$. Moreover, since they appear in pairs, there must be an even number of them. Hence, the characteristic polynomial must be of the form $$\chi(s) \propto (s-z_1)(s-z_1^*)\ldots(s-z_m)(s-z^*_m)$$ $$\chi(s) \propto (s^2-2\Re(z_1)s+\ell^2)\ldots(s^2-2\Re(z_m)s+\ell^2)$$ where $\ell$ is the common modulus of the roots which have a common ratio.

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  • $\begingroup$ You got the char equations. How about the matrices? $\endgroup$ – 1.. Oct 11 '13 at 20:45
  • $\begingroup$ @JAS But they are not proportional. Some have one root, some have $k$ roots, some have $\tfrac{1}{2}k$ pairs of complex conjugate roots if $k$ is even. $\endgroup$ – Fly by Night Oct 11 '13 at 20:47
  • $\begingroup$ Your first case is $r=1$ (just a repeat of second case). You also missed a case with $r=-1$. and if the root $a$ is generic, we will have all char equations propoertional to the second one. $\endgroup$ – 1.. Oct 11 '13 at 20:49
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    $\begingroup$ @JAS You asked "How should the char equation look like?" and that is what I tried to answer. $\endgroup$ – Fly by Night Oct 11 '13 at 20:49
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    $\begingroup$ @JAS No, the common ratio has modulus one, but the roots need not have modulus one. $\endgroup$ – Fly by Night Oct 11 '13 at 20:55

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