5. Factor out the unit-norm basis (and operator symmetry) tests once
all of the algebras pass.
-
-6. Implement spectral projector decomposition for EJA operators
- using jordan_form() or eigenmatrix_right(). I suppose we can
- ignore the problem of base rings for now and just let it crash
- if we're not using AA as our base field.
-
-7. Do we really need to orthonormalize the basis in a subalgebra?
- So long as we can decompose the operator (which is invariant
- under changes of basis), who cares?
"""
# The matrix method returns a polynomial in 'x' but want one in 't'.
return self.matrix().minimal_polynomial().change_variable_name('t')
+
+
+ def spectral_decomposition(self):
+ """
+ Return the spectral decomposition of this operator as a list of
+ (eigenvalue, orthogonal projector) pairs.
+
+ SETUP::
+
+ sage: from mjo.eja.eja_algebra import RealSymmetricEJA
+
+ EXAMPLES::
+
+ sage: J = RealSymmetricEJA(4,AA)
+ sage: x = sum(J.gens())
+ sage: A = x.subalgebra_generated_by(orthonormalize_basis=True)
+ sage: L0x = A(x).operator()
+ sage: Ps = [ P*l for (l,P) in L0x.spectral_decomposition() ]
+ sage: Ps[0] + Ps[1] == L0x
+ True
+
+ """
+ if not self.matrix().is_symmetric():
+ raise ValueError('algebra basis is not orthonormal')
+
+ D,P = self.matrix().jordan_form(subdivide=False,transformation=True)
+ eigenvalues = D.diagonal()
+ us = P.columns()
+ projectors = []
+ for i in range(len(us)):
+ # they won't be normalized, but they have to be
+ # for the spectral theorem to work.
+ us[i] = us[i]/us[i].norm()
+ mat = us[i].column()*us[i].row()
+ Pi = FiniteDimensionalEuclideanJordanAlgebraOperator(
+ self.domain(),
+ self.codomain(),
+ mat)
+ projectors.append(Pi)
+ return zip(eigenvalues, projectors)
sage: u = gram_schmidt(v)
sage: all( u_i.inner_product(u_i).sqrt() == 1 for u_i in u )
True
- sage: u[0].inner_product(u[1]) == 0
+ sage: bool(u[0].inner_product(u[1]) == 0)
True
- sage: u[0].inner_product(u[2]) == 0
+ sage: bool(u[0].inner_product(u[2]) == 0)
True
- sage: u[1].inner_product(u[2]) == 0
+ sage: bool(u[1].inner_product(u[2]) == 0)
True
TESTS:
# And now drop all zero vectors again if they were "orthogonalized out."
v = [ v_i for v_i in v if not v_i.is_zero() ]
- # Now pretend to normalize, building a new ring R that contains
- # all of the necessary square roots.
- norms_squared = [0]*len(v)
-
- for i in xrange(len(v)):
- norms_squared[i] = v[i].inner_product(v[i])
- ns = [norms_squared[i].numerator(), norms_squared[i].denominator()]
-
- # Do the numerator and denominator separately so that we
- # adjoin e.g. sqrt(2) and sqrt(3) instead of sqrt(2/3).
- for j in xrange(len(ns)):
- PR = PolynomialRing(R, 'z')
- z = PR.gen()
- p = z**2 - ns[j]
- if p.is_irreducible():
- R = NumberField(p,
- 'sqrt' + str(ns[j]),
- embedding=RLF(ns[j]).sqrt())
-
- # When we're done, we have to change every element's ring to the
- # extension that we wound up with, and then normalize it (which
- # should work, since "R" contains its norm now).
+ # Just normalize. If the algebra is missing the roots, we can't add
+ # them here because then our subalgebra would have a bigger field
+ # than the superalgebra.
for i in xrange(len(v)):
- v[i] = v[i].change_ring(R) / R(norms_squared[i]).sqrt()
+ v[i] = v[i] / v[i].norm()
return v