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Let $A,B$ be two matrices, $\rho$ be spectral radius, which is the top eigenvalue of a matrix. I discovered that $$\rho(AABABB)=\rho(ABAABB).$$ But I could not find the reason.

By the way, all I had tested were $2\times2$ positive matrices. And I know the fact $\rho(AB)=\rho(BA)$, but it seems that this property is not enough for this statement.

Fan
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  • When you say you discovered, do you meant you tried it on a few examples? – copper.hat Nov 10 '14 at 03:26
  • Yes. I programmed and found this result. May not be generally true although – Fan Nov 10 '14 at 03:27
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    If this is really true for all $2\times 2$-matrices $A,B$, then the same equality must be valid for the bottom eigenvalues (because replace $A$ and $B$ by $A^{-1}$ and $B^{-1}$ and replace some matrices by similars). Therefore, $AABABB$ and $ABAABB$ must have the same characteristic polynomial (for $2\times 2$-matrices). This should be easy to check? – darij grinberg Nov 10 '14 at 03:55
  • It is true for all $2\times2$ matrices $A$ and $B$ over $\Bbb{C}$. – Servaes Nov 10 '14 at 04:09
  • Is there a nice noncomputational proof of this? Using, say, the Cayley-Hamilton theorem? – darij grinberg Nov 10 '14 at 04:12
  • Interesting question. How did it arise? – user1551 Nov 10 '14 at 08:14
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    @user1551 I tried to find a quick method to tell if two sequences are cyclic permutation equivalent (for example, $12122$ and $22121$ are equivalent because we can take $22$ to the front). So I think of $\rho$ to be an invariant of cyclic permutation. But this example shows that it is not a good invariant. – Fan Nov 10 '14 at 08:34
  • Related: http://math.stackexchange.com/questions/29066/operatornametraababb-operatornametraabbab-for-2%C3%972-matrices – Myself Jan 03 '15 at 11:46

2 Answers2

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Your statement is true for all $2\times2$ matrices (positive or not). Recall that matrix trace of matrix product is invariant under cyclic permutation. Let $\renewcommand{\tr}{\operatorname{tr}}t=\tr(AB)=\tr(BA)$ and $d=\det(AB)=\det(BA)$. Then \begin{align} \renewcommand{\tr}{\operatorname{tr}} \tr(AABABB) &= \tr[(AB)^2 BA] = \tr[(tAB-dI)BA] = \tr(tAABB - dAB),\\ \tr(ABAABB) &= \tr[(BA)^2 AB] = \tr[(tBA-dI)AB] = \tr(tAABB - dAB). \end{align} Hence the two matrix products have identical traces. Obviously they have identical determinants too. So, by Cayley-Hamilton theorem, they have identical spectra and in turn identical spectral radii.

I guess the statement is not true for larger-sized matrices, but I don't have time to perform any computational test.

user1551
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  • Thanks for your answer. I tried $3\times 3$ and $4\times 4$ case, it seems they are not true. But interesting enough, if only positive matrices considered, all the eigenvalues of $AABABB$ and $ABAABB$ are pretty close. – Fan Nov 10 '14 at 08:29
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    @Fan Pretty close?! If $A$ and $B$ are Hermitian (positive or not), the two spectral radii should be identical to each other, because $$\rho(AABABB) = \rho((AABABB)^\ast) = \rho(BBABAA) = \rho(BB(ABAA)) = \rho((ABAA)BB)!!!$$ – user1551 Nov 10 '14 at 09:15
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Here is a less satisfactory negative answer using Octave:

octave:1> a=rand(3,3)
a =

   0.853184   0.968858   0.369978
   0.995402   0.194116   0.449798
   0.373102   0.045246   0.894742

octave:2> b=rand(3,3)
b =

   0.60655   0.11672   0.90867
   0.30900   0.89411   0.17607
   0.14202   0.94938   0.92741

octave:3> abs(eig(a*a*b*a*b*b))
ans =

   23.223839
    0.035014
    0.034103

octave:4> abs(eig(a*b*a*a*b*b))
ans =

   22.749277
    0.032287
    0.037755
copper.hat
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  • See, their eigenvalues are pretty close :) – Fan Nov 11 '14 at 05:35
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    This was just to illustrate that they are not equal. If I fiddle with the distribution (each entry in $a,b$ above is uniformly selected from $(0,1)$) I can get examples where the ratio of spectral radii is $>100$. – copper.hat Nov 11 '14 at 05:59