It’s interesting to think about this theorem (or really even just the second preparatory lemma) in terms of quantum mechanics. One of the first things a physicist learns is that if a set of operators commute with the Hamiltonian they are all simultaneously diagonalizable, meaning that there is a basis of states where each of the observables corresponding to those operators will be constant over time. But I suppose this result means that if there is a set of operators that may not commute but do form a solvable Lie algebra with the Hamiltonian, then there exists at least one stationary state in which all those observables are constant over time.
@iabervon Жыл бұрын
In both cases x-bar is well-defined because w is an eigenvector, which puts the image of possible differences between equivalent elements of the quotient space in C_w.
@NuclearCraftMod Жыл бұрын
Should we think of Engel and Lie’s theorems as end goals in intuiting the structure of nilpotent and solvable Lie algebras? It seems like the extent to which they are then used to prove further results is quite specialised.
@mudassirmudassir2944 Жыл бұрын
Sir please uplode more lectures I have needed
@pyrouscomments Жыл бұрын
there's something wrong with the first exercise: (1, 0, 0)^T is always a common eigenvector. You need non-triangular matrices in L for this to work.