An operator that commutes with all isometries is proportionnal to the identity matrix

Question

I have some trouble showing that a linear operator $L$ on some finite dimensional inner product space with is such that $K L=LK$ for all isometries $K$ of $V$ has necessarily the form $L=\lambda \cdot id_V$.

Answer 1

Pick any $v\in V$ with $|v|=1$. Then we can write $Lv=av+v'$ with $ v'\perp v$ (namely, $a=\langle v,Lv\rangle$). We can find $K$ with $Kv=v$ and $Kv'=-v'$. Then $$av-v'=K(av+v')=KLv=LKv=Lv=av+v'.$$ We conclude $v'=0$, i.e., $v$ is an eigenvector with eigenvalue $a$. Let $L'=L-aI$. Then $L'$ also commutes with isometries, and we have $L'v=0$. For any $u\in V$, we can find an isometry $K'$ with $K'u=|u|v$. Then $$ K'L'u=L'K'u=|u|L'v=0.$$ As $K'$ is injectve, we conclude $L'u=0$, i.e., $$Lu=\lambda u.$$

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Bonus question: Where dis I use $\dim V<\infty$?

Answer 2

Maybe there is no need for $V$ to be finite dimensional.

For $v\in V$ with $\|v\|=1$ let $\langle v\rangle^{\bot}$ be the orthogonal complement of $\langle v\rangle$. Since $\langle v \rangle$ is finite dimensional $V=\langle v\rangle \oplus \langle v\rangle^{\bot}$. By choosing isometry $K$ to be $id_{\langle v\rangle}\oplus -id_{\langle v\rangle^{\bot}}$ and using $Lv = \langle v ,Lv\rangle v + Lv - \langle v ,Lv\rangle v$ it follows that $v$ is an eigenvector of $L$. Since this can be done for any $v$ it follows that $L=\lambda \cdot id$.