For a matrix with eigenvalues the k-roots of unity, will we be sure to have it block-similar to a k size circulant generator matrix $\bf C_4$:

$${\bf C}_4 = \left[\begin{array}{cccc}0&1&0&0\\0&0&1&0\\0&0&0&1\\1&0&0&0\end{array}\right]$$

I have managed to do the opposite, randomizing 4x4 $\bf V$ transform matrix, and then calculated

$$\text{eig}({\bf VC_4V}^{-1})$$

giving me the eigenvalues (from Gnu Octave)


which seems very close to $[1,i,-1,-i]$ given double precision having been used.

Will this work in general and the other way around?

i.e. if we find matrix $\bf A$ has $$\lambda_k = \exp\left(\frac{-2\pi ik}{n}\right), \forall k \in[1,n]$$

Are we sure we can find a $\bf V$ so that : $\bf {A = VC_nV}^{-1}$

  • 2
    $\begingroup$ Hint: if two $A \sim diag(\lambda_1, \ldots, \lambda_n)$ and $B \sim diag(\lambda_1, \ldots, \lambda_n)$ then $A \sim B$. And if $A$ has for eigenvalues $\lambda_1, \ldots, \lambda_n$ then $A$ has $n$ distinct eigenvalues and $A \sim diag(\lambda_1, \ldots, \lambda_n)$$. $\endgroup$ – Delta-u Apr 28 '18 at 15:06
  • $\begingroup$ Omg, that is what I was thinking. But I thought in terms of matrices, i.e. $D={V_1AV_1}^{-1} = {V_2C_nV_2}^{-1}$ and then figure out if we can rearrange the Vs. $\endgroup$ – mathreadler Apr 28 '18 at 15:13
  • 1
    $\begingroup$ You can rearrange the $V$: $$V_2^{-1} V_1 A V_1^{-1} V_2=C_n$$ i.e: $$(V_2^{-1} V_1) A (V_2^{-1} V_1)^{-1}=C_n$$ $\endgroup$ – Delta-u Apr 28 '18 at 15:15
  • $\begingroup$ @Delta-u Yes as I hoped, dat is schweet. How extremely cool! Finding vectors which rotate around given some transformation without having to invent silly numbers that don't exist. $\endgroup$ – mathreadler Apr 28 '18 at 15:17
  • $\begingroup$ Feel free to make it an answer if you want to, I think it deserves an upvote. $\endgroup$ – mathreadler Apr 28 '18 at 15:19

If an $n \times n$ complex matrix has $n$ distinct eigenvalues $\lambda_1, \ldots, \lambda_n$, we know that it is similar to the diagonal matrix $\mathrm{diag}(\lambda_1, \ldots, \lambda_n)$. Hence any two $n \times n$ matrices with the distinct eigenvalues $\lambda_1, \ldots, \lambda_n$ will be similar. So your problem is equivalent to asking whether the $n \times n$ circulant matrix $C_n$ has eigenvalues given by $\mathrm{exp}(2 \pi i k / n)$ for $k \in \{1, \ldots, n\}$. And this should be easy to check, since the eigenvectors have a nice form as powers of $n$th roots of unity.


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