Day 2.2/168: if you could just as easily act on either, we can map the whole town by short rival streets
spectral theorems for self-adjoint / unitary operators on finite-dimensional vector spaces
if you could just as easily act on either, we can map the whole town by short rival streets
You: a self-adjoint map
Because we can work in the rich vocabulary, then realize we only ever needed simple words1, we know there’s some lengthy street2 you must scale. The street keeps to itself3, which polarizes4, so the rival hood5 keeps to itself and your tougher parts 往復同效 there too6. They keep squeezing into tougher & tougher parts of town7, discovering novel insular streets8, noting down their source code9, to exhaustion10. Add the original street’s source code11, and you’ve got a minimal map for the whole town.12
parallel case: if you’re distance-preserving, we can map the whole town by short rival streets
It’s the same proof, but in each case, we show T-invariance of orthogonal complement / “The street keeps to itself, which polarizes” differently.
In the case where you could just as easily act on either (self-adjoint map):
Take some rival hood member. How similar are the members after you act on the rival? … not at all, because you could just as easily act on either, and the originals keep to themselves, and they are rivals after all.
In the case where you’re distance-preserving (unitary map):
Take some rival hood member. Acting on the rival is like supplanting the original with its rewind.13 Write out your kryptonite, (it’s full kryptonite because you’re invertible so all roots are non-zero14), shift your roots to one side, factor yourself out; you get an expression for your rewind in terms of yourself. So the original keeps to itself when rewound; rewind + rival are full rivals; rival hood member keeps to itself when you act on it.
we can work over C => because C is algebraically closed, characteristic polynomial has a root => ah, these roots / eigenvalues are real (because T self-adjoint)
some non-zero real-eigenvalued eigenvector
is T-invariant
makes its orthogonal complement you-invariant. by prior lemma
street’s orthogonal complement
your restriction to that complement is self-adjoint and well-defined
we perform induction on n := dim(V)
finding new T-invariant eigenvectors
just take a cutting/clipping, the normalization. (i’m thinking ‘short street’ := normalized eigenvector; ‘long street’ := span of eigenvector)
induction ends when you hit a one-dimensional subspace
add the first eigenvector, normalized
you have an orthonormal basis for V consisting of your eigenvectors
T-preimage
a_0 != 0