Let $A$ be an $n\times n$ invertible complex matrix such that $A^7 = A^*$ (where $*$ denotes conjugate transpose). Show that $A^8 = I$.

Here are my thoughts so far:

  • I was able to show that all the eigenvalues of $A$ satisfy $\lambda^8 = 1$.

  • I tried writing $A$ in Jordan canonical form $A = PJP^{-1}$ so that $A^* = (P^{-1})^*J^*P^*$ and $A^7 = PJ^7P^{-1}$. I was hoping to conclude $J^* = J^7$ but this isn't necessarily true as they may not be Jordan matrices; they would be if $J$ was diagonal, but if I knew that $J$ was diagonal, I would be done.

  • As $A^{49} = (A^7)^7 = (A^*)^7 = (A^7)^* = (A^*)^* = A$, $A^{48} = I$ so the minimal polynomial of $A$ divides $x^{48}-1$; note that $x^8 - 1$ is a factor of $x^{48}-1$.

  • If $A^8 = PJP^{-1}$ is the Jordan normal form of $A^8$, then $J = I + N$ and $J^6 = I$ by the previous point so $(I+N)^6 = I$ but I can't directly deduce from this equality that $N = 0$.

Any hints are very much appreciated.


Since $A^7=A^*$ we have $A^*A=AA^*$ so $A$ is a normal matrix. Thus $A$ is diagonalisable, and since all its eigenvalues are roots of $X^8-1=0$ we conclude that $A^8=I$.

Edit: Indeed, consider $x\ne0$ an eigenvector of $A$ corresponding to the eigenvalue $\lambda$. We have $$ \bar{\lambda}\Vert x\Vert^2=\langle x,Ax\rangle=\langle A^*x,x\rangle =\langle A^7x,x\rangle=\lambda^7\Vert x\Vert^2 $$ Thus $\lambda^7=\bar{\lambda}$. In particular, $|\lambda|^7=|\lambda|$, and since $\lambda\ne0$ because $A$ is invertible we conclude that $|\lambda|=1$, and consequently $\lambda^7=\bar{\lambda}=1/\lambda$ or $\lambda^8=1$.

  • 1
    $\begingroup$ can you please explain why the eigenvalues of $A$ are eighth root of unity? $\endgroup$ – GA316 Aug 15 '14 at 7:06
  • $\begingroup$ @GA316, I explained this. $\endgroup$ – Omran Kouba Aug 15 '14 at 7:26
  • $\begingroup$ thanks a lot. nice explaination $\endgroup$ – GA316 Aug 15 '14 at 8:48
  • $\begingroup$ Very nice. The edit explains that the same argument would work with any integer exponent $m$ instead of $7$ (the conclusion then being $A^{m+1}=I$), except for $m=1$ (in which case the conclusion $A^2=I$ would actually be false in general). $\endgroup$ – Marc van Leeuwen Aug 16 '14 at 10:40

In regards to your last point, it suffices to work on one Jordan block at a time. Let $N$ be the nilpotent part of an $e\times e$ Jordan block. Examine $\{I,N,N^2,\cdots,N^{e-1}\}$, see what the elements look like explicitly. In particular, consider linear dependence or independence.


I use $A^\dagger$ for the Hermitian adjoint, or conjugate transpose, of $A$.

We have

$A^7 = A^\dagger; \tag{1}$

this implies that $A$ is normal, or $AA^\dagger = A^\dagger A$:

$AA^\dagger = AA^7 = A^7A = A^\dagger A; \tag{2}$

since $A$ is normal, there is a unitary matrix $U$ which diagonalizes $A$, thus:

$U^\dagger A U = \Lambda, \tag{3}$

where $\Lambda$ is a diagonal matrix whose diagonal is comprised of the eigenvalues of $A$; see this widipedia entry on normal matrices. Furthermore,

$U^\dagger A^\dagger U = (U^\dagger A U)^\dagger = \Lambda^\dagger, \tag{4}$

and $\Lambda$ being diagonal, we have $\Lambda^\dagger = \Lambda^\ast$, the matrix of complex conjugates of $\Lambda$; this illustrates the reason I chose ${}^\dagger$ for adjoint; I needed to reserve ${}^\ast$ for (elementwise) conjugation. In any event, (4) thus yields

$U^\dagger A^\dagger U = \Lambda^\dagger = \Lambda^\ast. \tag{5}$

Now (1) implies

$(U^\dagger A U)^7 = U^\dagger A^7 U = U^\dagger A^\dagger U, \tag{6}$

so by (3) and (4)

$\Lambda^7 = \Lambda^\ast; \tag{7}$

this shows that for any diagonal entry $\lambda$ of $\Lambda$ we must have

$\lambda^7 = \lambda^\ast, \tag{8}$


$\vert \lambda \vert^7 = \vert \lambda^7 \vert = \vert \lambda^\ast \vert = \vert \lambda \vert; \tag{9}$

and since $A$ is invertible, $\lambda \ne 0$, so

$\vert \lambda \vert^6 = 1, \tag{10}$

whence we must have

$\vert \lambda \vert = 1, \tag{11}$

$\vert \lambda \vert$ being positive real. All the $\lambda$ are therefore unimodular, yielding

$\lambda^\ast = \lambda^{-1}; \tag{12}$

Applying (12) to (8) gives

$\lambda^8 = 1, \tag{13}$


$\Lambda^8 = I. \tag{14}$

Finally, (3) implies

$A^8 = (U \Lambda U^\dagger)^8 = U I U^\dagger = I, \tag{15}$

as per request. QED.

Nota Bene: There seems to be no magic in the hypothesis $A^7 = A^\dagger$, other than the general magic of mathematics: apparently $A^n = A^\dagger$ implies $A^{n + 1} = I$ for any integer $n \ge 2$ and invertible $A$. End: Nota Bene.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.