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Let L be a finite-dimensional complex semisimple lie algebra, then ad(L)=Der(L). (Der is short for derivation). In order to show that ad(L)=Der(L), the proof I followed proves that that the prependicular space to ad(L) is zero. (Perpendicular with respect to killing form of Der(L).) I understand the proof of this. But why is this condition sufficient to conclude ad(L) = Der(L)

If we knew that Der(L) is the direct sum of ad(L) and P:=Perp(ad(L)) then we would be done. And non-degeneracy of the killing form restricted to ad(L) tells us that it is sufficient to show Dim(Der(L)) = Dim(ad(L)) + Dim(P). But I am failing to see how we have that.

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This is a homework problem from a first course in Lie algebras, so I think it's not appropriate here. – Robert Bryant Jan 16 '13 at 20:15
Whether or not it's homework-related, I agree the question is far from research level (try It's a standard theorem found in many books with similar proofs, attributed by Jacobson to Zassenhaus. Not conceptually transparent but a fairly elementary consequence of nondegeneracy of the Killing form. – Jim Humphreys Jan 17 '13 at 1:30

If $U$ is a vector subspace of a finite-dimensional vector space $V$ such that the perpendicular space of $U$ with respect to some symmetric bilinear form on $V$ (which does not need to be nondegenerate) is zero, then $U=V$. This is very easy to check (assume the contrary, then notice that $\dim U < \dim V$, and thus the perpendicular space of $U$ has codimension $\leq \dim U$, which means it cannot be zero). You don't need direct sum decompositions.

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Where have you got Dim(V/Perp(U)) ≤ Dim(U). – rustyracketman Jan 16 '13 at 18:03
The perpendicular space of $U$ is given by the condition that it be perpendicular to all elements in a (fixed) basis of $U$. These are $\dim U$ linear equations, and thus the vector space of their joint solutions has codimension $\leq \dim U$. – darij grinberg Jan 16 '13 at 19:26

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