Taking the semidirect product of this 3-dimensional Lie algebra by the " p "-dimensional representation ( considered as an abelian Lie algebra ) gives a solvable Lie algebra whose derived algebra is not nilpotent.
2.
In fact it is enough to show there is a vector annihilated by, which follows by induction on, since the derived algebra annihilates a non-zero subspace of vectors on which and act with the same hypotheses.
3.
Because a weight is a map to a commutative ring, the map factors through the abelianization of the algebra \ mathcal { A } equivalently, it vanishes on the derived algebra in terms of matrices, if v is a common eigenvector of operators T and U, then T U v = U T v ( because in both cases it is just multiplication by scalars ), so common eigenvectors of an algebra must be in the set on which the algebra acts commutatively ( which is annihilated by the derived algebra ).
4.
Because a weight is a map to a commutative ring, the map factors through the abelianization of the algebra \ mathcal { A } equivalently, it vanishes on the derived algebra in terms of matrices, if v is a common eigenvector of operators T and U, then T U v = U T v ( because in both cases it is just multiplication by scalars ), so common eigenvectors of an algebra must be in the set on which the algebra acts commutatively ( which is annihilated by the derived algebra ).
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