(1). Let $sed{A_al}$ be a family of mutually commuting operators. Then, there exists a common Schur basis for $sed{A_al}$. In other words, there exists a unitary $Q$ such that $Q^*A_al Q$ is upper triangular for all $al$.
(2). Let $sed{A_al}$ be a family of mutually commuting normal operators. Then, there exists a unitary $Q$ such that $Q^*A_al Q$ is diagonal for all $al$.
Solution.
(1). We may assume $A_al$ is not the multiplier of the identity operator (otherwise, we could just delete it). We prove by induction on the dimension $n$ of the vector space $scrH$ we consider. If $n=1$, then it is obvious true. Suppose the conclusion holds for vector spaces with dimension $leq n-1$. To prove the statements for the case $dim scrH=n$, we need only to prove that there exists an one-dimensional subspace that is $A_al$-invariant for each $al$. In fact, $$eex ea &quad sex{a{cc} 0&b\ 0&B ea}sex{a{cc} 0&c\ 0&C ea}=sex{a{cc} 0&c\ 0&C ea}sex{a{cc} 0&b\ 0&B ea}\ & a sex{a{cc} 0&bC\ 0&BC ea}=sex{a{cc} 0&cB\ 0&CB ea}\ & a BC=CB. eea eeex$$ Fix a $eta$, suppose $lm$ is an eigenvalue of $A_eta$, then $$ex W=sed{xinscrH; A_eta x=lm x} eex$$ is $A_al$-invariant. Indeed, $$ex A_eta A_al x=A_al A_eta x=lm A_al x. eex$$ Thus, $W eq scrH$ (by the fact that $A_eta$ is not the multiplier of the identity operator), and $$ex dim W<dim scrH. eex$$ Also, $A_al$ may be viewed as a commuting operator on $W$, and the induction hypothesis may be invoked to deduce that there exists a orthonomal basis $x_1,cdots,x_k$ of $W$ such that $$ex A_al(x_1,cdots,x_k)=(x_1,cdots,x_k)sex{a{ccc} *&&*\ &ddots&\ 0&&* ea}. eex$$ The subspace spanned by $x_1$ is then one-dimensional, and is $A_al$-invariant for each $al$.
(2). By (1), $exists$ unitary $Q$ such that $A=QU_al Q^*$ for some upper triangular $U_al$. Since $A_al$ is normal, we have $U_al^*U_al=U_al U_al^*$. By comparing the diagonal entries, we see readily that $U_al$ is diagonal, as desired.