Connection between permutations and rank–nullity?

Question

A fact about permutations on $n$ letters is that if for $\sigma\in S_n$, $O(\sigma)$ denotes the number of orbits/cycles in $\sigma$, and $T(\sigma)$ denotes the minimum number of transpositions needed to write $\sigma$, then $$O(\sigma)+T(\sigma)=n.$$ (The proof comes down to each cycle of length $m$ in $\sigma$ (even the fixed points) needing $m-1$ transpositions to write.) However, $S_n$ also naturally embeds into the endomorphisms of a $n$-dimensional vector space by permuting any chosen basis. Thus, the result about $O$ and $T$ looks a bit like a rank–nullity type result. My question then is: is there a "canonical" way to identify the symmetric group with some class of linear operators, such that $O$ is the rank and $T$ the nullity, or vice versa? ("Canonical" can be taken somewhat loosely, but the permutation data of each $\sigma$ should meaningfully inform the corresponding operator.) Unfortunately, the most immediate candidate, the permutation representation, fails as it has full rank and 0 nullity.

Answer 1

Eigenvectors of the permutation matrix corresponding to the eigenvalue $1$ can be obtained from cycles of the permutation — each cycle gives an eigenvector, which is the sum of basis vectors corresponding to the elements of the cycle, and these eigenvectors span the eigenspace of $1$. So, if $P$ is the permutation matrix, then $O$ is the nullity of $P - \mathbf{1}$, and $T$ is its rank.