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I want to start a collection of statements of the following form:

If a linear operator $T:V\to V$ on the vector space $V$ commutes with all elements of the set $\Phi$, i.e. $U T = TU$ for all $U\in \Phi$, then $T$ is of the form ...

Of course if this statement is true for a small $\Phi$, it is also true for all larger $\Phi$. There have been many questions, which ask for a proof that if $T$ commutes with all linear operators it is a scalar multiple of the identity (e.g. 2011,2012a,2012b,2013). But these results are corollaries of much smaller $\Phi$ e.g.

If $\dim(V)<\infty$ and $T$ thus a matrix, we furthermore have:

Please only post one result per answer (the statement as a headline with a proof below)

Felix Benning
  • 2,889

2 Answers2

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$\dim(V)<\infty$, $\Phi$ the set of Permutations

then

$$ T=\lambda \mathbb{I} + \mu \mathbf{1} \mathbf{1}^T $$ for $\mathbf{1}=(1,\dots,1)^T$ and some scalars $\lambda, \mu\in \mathbb R$

Proof (copying Ben Grossmann): Let $P$ be the permutation which swaps $e_i$ with $e_j$ i.e. $P e_i = e_j$, $Pe_j = e_i$ and $Pe_k = e_k$ for $k\neq i,j$. Since $T$ commutes with $P$, i.e. $PT = TP$, we have $T = P^T T P$. Thus

$$ \begin{aligned} T_{ij} &= e_i^T T e_j = e_i^T P^T T P e_j = T_{ji} \\ T_{ii} &= T_{jj} \\ T_{ik} &= T_{jk} && k\neq i,j \\ T_{kj} &= T_{ki} && k\neq i,j \end{aligned} $$

By choosing various $P$, the equations above hold for all distinct $i\neq j$.

The first two equations lines are enough for us to deduce that $T$ is symmetric with a constant diagonal. With the third/fourth line, we see that in the $k$-th row/column, the off-diagonal entries are all equal. Thus, all off-diagonal entries of $T$ are equal.

Thus, we've deduced that has the form $$T=\lambda \mathbb{I} + \mu \mathbf{1} \mathbf{1}^T$$ where $\mathbf{1}=(1,\dots,1)^T$.

Felix Benning
  • 2,889
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$\dim(V)<\infty$, $\Phi$ the image of an irreducible representation $(\rho, V)$ of a group $G$

then $$ T = \lambda \mathbb I $$ for some $\lambda \in K$ for an algebraically close field $K$ (e.g. $\mathbb C$).

Proof: Schur's Lemma since $T$ is a $G$-linear map w.r.t. the group $G$, as it commutes with all its representations.

The problem with this result is that checking whether $\Phi$ is in fact the image of an irreducible representation of the group $G$ is non-trivial in general. But a unitary representations (which maps into the set of unitary matrices) can be (uniquely?) decomposed into a direct sum of irreducible representations by Maschke's Theorem.

Example (Permutations): Since permutation matrices are orthogonal matrices Maschke's theorem is applicable. Since the vector $\mathbf{1}=(1,\dots, 1)^T$ forms an invariant subspace, the representation can be decomposed into representations on $V_1:=\mathrm{span}(\mathbf{1})$ and $V_1^\perp$.

And $T$ can be similarly decomposed into $T_{|V_1}$ and $ T_{|V_1^\perp}$, which can both be expressed as scalar multiples of the identity but not necessarily the same scale. This leads to the sum representation of the corresponding answer, where we note that $\mathbf{1}\mathbf{1}^T$ is the projection matrix onto $V_1$ up to scaling.

Felix Benning
  • 2,889