0
$\begingroup$

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.

$\endgroup$
0
$\begingroup$

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 $k>1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.