Prove that every permutation matrix satisfies its characteristic polynomial.

Question

Let $P$ is a permutation matrix which represents the permutation $\sigma\in S_{n}$. Let $\sigma_{1}$,$\sigma_{2}$,...$\sigma_{k}$ denote the disjoint permutations in the cycle form of $\sigma$. Let $P_{i}$ and $c_{i}$ represents the permutation matrix corresponding to the permutations $\sigma_{i}$ and the cycle lengths of $\sigma_{i}$ respectively. Prove that P satisfies the equation (its characteristic polynomial) $$\prod_{i=1}^{k}(P^{c_{i}}-I) =0$$

I know the following facts:

1. For disjoint permutation matrices $P_{i},P_{j}$ we have $(P_{j}-I)(P_{i}-I) = 0$.

2. For disjoint permutation matrices $P_{1},P_{2},\cdots,P_{k}$ we have $$\prod_{i=1}^{k}P_{i} = \sum_{i=1}^{k}P_{i}-(k-1)I.$$

3. If P and Q are disjoint permutation matrices so are $P^{m}$ and $Q^{n}$ , $\forall m,n\in \Bbb N$.

4. If P and Q are disjoint permutation matrices they commute.

5. If P is a single-cycled permutation matrix with cycle length $k$ then $P^{k}=I$.

6. Combining the fact $2$ and $3$ for disjoint permutation matrices $P_{1},P_{2},\cdots,P_{k}$ we also have $$\prod_{i=1}^{k}P_{i}^{m} = \sum_{i=1}^{k}P_{i}^{m}-(k-1)I, \forall m\in \Bbb N.$$

7. Combining the fact $1$ and $3$ we have $(P_{j}^{n}-I)(P_{i}^{m}-I) = 0$ for any $n,m \in \Bbb N.$

MY TRY: I tried with an case where $P$ breaks in two single-cycled disjoint permutations $Q$ and $R$ with cycle lengths $m,n$ respectively. We need to prove that $$(P^{n}-I)(P^{m}-I) = \Bigr((QR)^{n}-I\Bigl)\Bigr((QR)^{m}-I\Bigl) = 0$$
Using fact $4$ and $5$ $$\Bigr((QR)^{n}-I\Bigl)\Bigr((QR)^{m}-I\Bigl)=\Bigr(Q^{n}R^{n}-I\Bigl)\Bigr(Q^{m}R^{m}-I\Bigl) = \Bigr(R^{n}-I\Bigl)\Bigr(Q^{m}-I\Bigl)$$ The fact $7$ as stated above states that it vanishes. But it gets more calculative when P breaks in $3$ disjoint single cycled permutations. Also, generalisation would need more calculations.

I do not know Cayley hamilton theorem. I am new to group theory. Please ask for clarifications if something is not clear. Any hint would be a great help.

Answer 1

I think this is much easier than you are making it out to be. By relabelling, you may assume that the permutation is $$(1,\dots,c_1)(c_1+1,\dots,c_1+c_2)...$$ The permutation matrix of this product of disjoint cycles is a block-diagonal matrix, with the blocks being the permutation matrices of each cycle.

Products and sums of block-diagonal matrices are block-diagonal, obtained by taking the products and sums of each block. Thus a block-diagonal matrix satisfies a polynomial if and only if each block of it does.

Block $i$ certainly satisfies the polynomial $P^{c_i}-I$, your fact 5. Thus this is the zero matrix, and the product of this with anything else is zero, in particular your polynomial is the zero matrix on the $i$th block. Thus your matrix is zero.

Answer 2

This is just an even more elementary version of David Craven's answer.

For each cycle $(i_1, \ldots, i_c)$ of $\sigma$, note that $P^c e_{i_j} = e_{i_j}$, so $P^c - I$ vanishes on $\mathrm{Span}(e_{i_1}, \ldots, e_{i_c})$. Thus $\prod_{i=1}^k (P^{c_i} - I)$ vanishes on this span as well, hence also on the span of these spans, which is the whole space.

The only fact of note we've used is, for every j, $$ \prod_{i=1}^k (P^{c_i} - I) = \left(\prod_{\substack{i=1 \\ i \neq j}}^k (P^{c_i} - I)\right) (P^{c_j} - I), $$ which is of course immediate, e.g. $$(P^a - I)(P^b - I) = P^{a+b} - P^a - P^b + I = (P^b - I)(P^a - I).$$

Answer 3

As, Disjoint permutations are commutative $$\prod_{i=1}^{k}(P^{c_{i}}-I) =\prod_{i=1}^{k}\Biggl(\biggl(\prod_{j=1}^{k}P_{j}\biggr)^{c_{i}}-I\Biggl) = \prod_{i=1}^{k}\Biggl(\prod_{j=1}^{k}P_{j}^{c_{i}}-I\Biggl)$$ We call the above equation as $[1]$. Your fact $6$ tells us that

The product of $k$ disjoint permutation matrices equals Identity matrix subtracted $k-1$ times from the sum of those permutation matrices

Also, as $P_{i}^{c_{i}}=I$. Using these both, the product term in the rightmost expression in [$1$] $$\prod_{j=1}^{k}P_{j}^{c_{i}}=\sum_{j=1}^{k}P_{j}^{c_{i}}-(k-1)I=\sum_{j\neq i}^{k}P_{j}^{c_{i}}-kI$$ Plugging it back in the rightmost term in the equation [$1$], equation [$1$] becomes $$\prod_{i=1}^{k}(P^{c_{i}}-I) =\prod_{i=1}^{k}\Biggl(\biggr(\sum_{j\neq i}^{k}P_{j}^{c_{i}}-kI\biggl)-I\Biggl)=\prod_{i=1}^{k}\Biggl(\sum_{j\neq i}^{k}(P_{j}^{c_{i}}-I)\Biggl)$$
Let's now handle the product on the right. Before that, let $A_j$ denote the matrices of the kind $P_{j}^{a}-I$ for any $a$ . First, we should understand the following two properties for $i,j\in \Bbb N$

$1$. $A_jA_i = 0$ for $j\neq i$ (immediate from the fact $7$)

$2$. $A_jA_i=A_iA_j$

Proof of 2: As $P_{j}^a$ and $P_{i}^b$ are disjoint they are commutative so, $(P_{j}^a-I)(P_{i}^b-I)=P_{j}^aP_{i}^b-P_{i}^b-P_{j}^a+I=P_{i}^aP_{j}^b-P_{j}^a-P_{i}^b+I=(P_{i}^b-I)(P_{j}^a-I)$

Coming back to product, we are left with the product of $k$ terms $$(A_2+A_3...+A_k)(A_1+A_3...+A_k)...(A_1+A_2...+A_{k-1})$$ The resulting expression will be a "polynomial of degree $k$ in $A$". But, We can't have a term like $A_i^k$ in the "polynomial" for any $i$, as the $i_{th}$ bracket in this product does not have $A_i$.

Hence, we must have terms in the polynomial involving product of at least two different $A_i's$. But by commutativity of $A_i$'s we can take any such pair of cross terms together. So, the polynomial is just the sum of terms of the form $MA_iA_j$ for some $i\neq j$ and $M$ is a matrix of "degree" $(k-2)$ in $A$. But the property $1$ says that every such term is a zero matrix. Hence, we conclude that the product which resulted in such a polynomial is zero matrix. Leaving us with,
$$\prod_{i=1}^{k}(P^{c_{i}}-I)=0.$$