How do I show that a linear operator commuting with all isometries is a scalar multiple of the identity?

Question

I'm trying to show that an operator $L$ on a finite dimensional inner product space that commutes with all isometries on $V$ must necessarily be a scalar multiple of the identity. I know a proof of the fact that only scalar multiples of the identity commute with all operators on any finite dimensional space. But I can't prove this. I'd appreciate some help. Thanks.

Answer 1

The key is to make sure you're looking at enough isometries.

Let $\{e_1,\dots,e_n\}$ denote the canonical (orthonormal) basis. Let $P$ be the isometry that transposes two basis elements, so $Pe_i = e_{j}$ and $P e_j = e_i$, and $Pe_k = e_k$ for $k \neq i,j$.

$L$ commutes with $P$, so $LP = PL$, so $L = PLP^T$. This means that the matrix of $L$ satisfies $$ L_{ij} = L_{ji} \\ L_{ii} = L_{jj} \\ L_{iq} = L_{jq} \quad q \neq i,j\\ L_{qi} = L_{qj} \quad q \neq i,j\\ $$ and the above holds for any choice of distinct $i$ and $j$.

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

Thus, we've deduced that $L$ has the form $$ L = \lambda I + \mu xx^T $$ where $x = (1,1,\dots,1)^T$.

Now, it suffices to note that $L$ commutes with the isometry $Q$ defined by $Qe_1 = -e_1$, $Qe_k = e_k$ for $k>1$.