Let $V$ be a finite dimensional vector space over $\Bbb{C}$ and $\mathcal{F}\subset \text{End}(V)$ be a commuting family of diagonalizable operators on $V$ i.e. $\forall T\in \mathcal{F}$, $T$ is diagonalizable and $\forall T_1,T_2\in\mathcal{F},\ T_1T_2=T_2T_1$.
A function $r:\mathcal{F}\to\Bbb{C}$ is said to be a root of $\mathcal{F}$ if $\exists v\ne 0$ such that $T(v)=r(T)v\ \forall T\in\mathcal{F}$.
For A function $r:\mathcal{F}\to\Bbb{C}$, define $V(r):=\{v\in V|\ T(v)=r(T)v\ \forall T\in\mathcal{F}\}$. With this notation, $r$ is a root iff $V(r)\ne 0$.
We are asked to prove-
Keep $\mathcal{F}$ as above. Then
- There are finitely many roots of $\mathcal{F}$ say $r_1,r_2,\ldots,r_k$
- $V=V(r_1)\oplus V(r_2)\oplus\cdots \oplus V(r_k)$
I know this result holds when $V$ is finite dimensional complex inner product space and $\mathcal{F}$ is a commuting family of normal operators on $V$.
In this case, I only know (from Hoffman Kunze Linear Algebra book) that $\mathcal{F}$ is simultaneously diagonalizable i.e. there exists a basis of $V$ with respect to which all elements of $\mathcal{F}$ is diagonal matrix. But I don't know prove this statement from the simultaneously diagonalizable property of $\mathcal{F}$. Can anyone give me any idea how to prove this?
Thanks for help in advance.