In HP, do Wizards from different nationalities have different names for spells, and if so do they have to pronounce them as such during the casting?  It seems that more veteran wizards don't seem to need to pronounce many of the spells they are casting, and whilst I am sure this has been explained somewhere, I imagine it has something to do with loading the spell with meaning/empowerment.  However, younger wizards DO seem to need to say the words of the spell they are casting, and if so how is this affected by different languages? Would a mute non-muggle be unable to qualify for wizardry, for example?
The spells that we know about in the Harry Potter world all seem to be either in Latin, derived from Latin, or resembling Latin.  This makes sense for the European World, but there were likely witches and wizards in Africa, Asia, and the Americas.  These cultures would have developed independently, and the magical people in them would likely have still found ways to cast their spells.  Do cultures with languages that did not derive from Latin use the same words and phrases that are related to Latin for their spells?  Or did they have other phrases based on the languages in their cultures?
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The converse of Schur's Lemma in finite dimensional vector spaces is:     Let $V$ be a finite dimensional vector space over the complex numbers. Let $S$ be a set of endomorphisms of $V$ and assume that every endomorphism $A$ of $V$ such that $$AB=BA\text{ for all }B\in S$$ is of the form $\lambda I, \ \lambda\in\mathbb{C}$. Then $V$ is a simple $S$-space.   (*an endomorphism of $V$ is a linear operator from $V$ to $V$  *$V$ is a simple $S$-space if the only $S$-invariant subspaces of $V$ are $V$ itself and the zero subspace)  What I've tried so far:  Assume to the contrary that $V$ is not a simple $S-$space. Then, there exists a subspace $W$ of $V$ such that $W\not =\{0\}$, $W\not= V$ and $W$ is $S$-invariant. Let $W'=V\setminus W$. Then, since $V$ is finite, we can easily prove that $V=W\oplus W'$ (this is my attempt to translate Maschke's theorem in vector spaces). Consequently, for every $v\in V$, there exists unique $w\in W$ and $w'\in W'$, such that $v=w+w'$. Define the projection $P:V\rightarrow V$ by $Pv=w$ for every $v\in V$.   What is left to prove is that $PB=BP$ for all $B\in S$. Then $P$ is clearly not a scalar and thus we have the contradiction we are looking for.  There is a difficulty in showing that $PB=BP$ for all $B\in S$, because $W'$ might be $S$-invariant or not. More precisely:   Let $v\in V=W\oplus W'$ and $v\not =0_V$.  If $v\in W$ then $Bv\in W$ because $W$ is $S$-invariant, therefore: $PBv=Bv$ and $BPv=Bv$.  If $v\in W'$ then $Pv=0$, hence $BPv=B\cdot 0=0$, and  (i) if $Bv\in W'$ then $PBv=0$.  (ii) if $Bv\in W$ then $PBv=Bv$ and this is where the problem occurs.  Any hints, ideas or counterexamples would be very helpful. Complete userChrome.css specification in Firefox Zero integral of measurable $f$ on every interval implies $f=0$? 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