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Let $P_n=\begin{bmatrix} 1&0&0&0&0&\cdots&0\\0&0&1&0&0&\cdots&0\\0&0&0&0&1&\cdots&0\\\vdots\end{bmatrix}_{n\times n}$ be a permutation matrix where $n$ is odd i.e. if the rows and columns of $P_n$ are numbered $0$ to $n-1$, then the column position of $1$ in the $i^{th}$ row of $P_n$ is $2i\mod n$.

I am trying to find the eigenvalues of $P_n$?

I am looking at an answer here that says the eigenvalues of a permutation matrix with cycles of length $n_1,n_2,\cdots,n_r$ are $ \bigoplus_{j=1}^r\{e^{\ell 2\pi i / n_j}:\ \ell=0,\ldots,n_j-1\} $. However, I am unable to see what the cycles are for a general $n$ in my case. For example, for $n=5$, the cycles are $(0)(1243)$ but for $n=7$, the cycles are $(0)(124)(653)$. So I am not able to find the eigenvalues of $P_n$ for an arbitrary $n$. Any ideas?

1 Answers1

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I will prove the claim from Jaap's comment. As a consequence, we can deduce that all eigenvalues of $P_n$ will be of the form $\exp(2 \pi i \cdot k/m)$, where $m$ is the order of $2$ modulo $n$. That said, we don't necessarily know what the multiplicity of these eigenvalues will be.

Claim: The length of the longest cycle is the multiplicative order of $2$ mod $n$. The other cycle lengths divide that.

Proof: First, let $k$ denote the smallest positive integer for which $P_n^k = I$. Let $m$ denote the order of $2$ modulo $n$. I claim that $k = m$.

To see this, note that $P_n^m = I$ (why?). It follows that $k \mid m$ (why?). Conversely, the fact that $P_n^k = I$ means that $2^k \cdot i \equiv i \pmod n$ for all $i$. Since this holds for $i=1$, we see that $2^k \equiv 1 \pmod n$. Conclude that $m \mid k$. With that, we have $m = k$ as desired.

Now, decompose the permutation associated with $P_n$ into disjoint cycles. The fact that $P_n^m = I$ means that each cycle has an order that divides $m$. On the other hand, we see from the above claim that $m$ is the length of the cycle containing $i = 1$. So indeed, $m$ is the length of the longest cycle and the length of each cycle divides $m$.

Further observations of interest:

  • There will always be a cycle $(0)$ and associated eigenvalue $1$.
  • By Euler's theorem, we will always have $m \mid \varphi(n)$, where $\varphi$ denotes Euler's totient function.
  • For an arbitrary $i$, we have $$ 2^k i \equiv i \pmod n \iff n \mid (2^k - 1)i $$

Why $P_n^m = I$: note that $P_n^j$ is the matrix where the column position of $1$ in the $i$th row is $2^j \cdot i \pmod n$. The fact that $2^m = 1$ means that $P_n^m$ has a $1$ in the $i$th position of the $i$th row for each $i$, which is to say that $P_n^m$ is the identity matrix.

Why $k \mid m$: suppose to the contrary that $k \nmid m$. Write $m = q\cdot k + r$ (for some integers $q,r$ with $0< r < k$). We note that $$ P_n^r = P_n^{m - q \cdot k} = P_n^m \cdot (P_n^{k})^{-q} = I \cdot I = I, $$ contradicting our definition of $k$ as the smallest positive integer for which $P_n^k = I$

Ben Grossmann
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  • Thanks. Sorry, I don't see why $P^m=I$ and why that implies $k|m$? Out of touch with linear algebra. –  May 11 '21 at 19:04
  • @manifolded At this point you're more in the domain of abstract algebra than linear algebra. I could write out a full explanation if you like, but I think that you'll find it to be a worthwhile exercise to prove it for yourself. – Ben Grossmann May 11 '21 at 19:07
  • I see, thanks. I don't understand why we are looking at multiplicative order of $2$ (mod $n$) and not of any other number in $\mathbb{Z}_n$ that is coprime with $n$. Then, I don't understand how that relates to the order of $P$. It would be great if you can add an explanation. –  May 11 '21 at 19:21
  • @manifolded We're looking at the order of $2$ because the permutation is defined via multiplication by $2$. I'll add more when I get the chance. – Ben Grossmann May 11 '21 at 19:23
  • Last question, I understand now $k=m$. How does that imply $m$ is the length of the cycle containing $i=1$. Also, why is it important to know that the longest cycle contains $i=1$? Thanks. –  May 11 '21 at 19:57
  • @manifolded See my latest edit. For your latest questions, $P_n$ is the matrix associated with the permutation that takes $i$ to $2i$ (modulo $n$) or, depending on how you associate permutation matrices to permutations, possibly the inverse of this permutation. In either case, the fact that applying the permutation $m$ times maps $1$ back to $1$ means that multiplying by $2$ (or $2^{-1}$) that many times is the same as multiplying by $1$. It wasn't important that the longest cycle contains $i=1$, but it was important to establish that there is a longest cycle with length $m$. – Ben Grossmann May 11 '21 at 20:03
  • Perfect, understood! –  May 11 '21 at 20:08