"Symplecetification of vector fields is an isomorphic map of the Lie algebra of contact vector fields onto the Lie algebra of all locally hamiltonian vector fields with hamiltonians which are homogeneous of degree 1"
@TedShifrin someone had me prove that I can have $y=Px$ such that $\langle Ax,x \rangle = \|y\|^2$ and $\langle Bx, x\rangle = \langle y,Dy \rangle$ where $D$ is diagonal. that's quite cool.
@TedShifrin given $A$ positive-definite symmetric, $B$ symmetric, prove that there is an invertible $P$ such that $\langle Ax, x \rangle = \|Px\|^2$ and $\langle Bx, x \rangle = \langle Px,DPx \rangle$ for all $x$, where $D$ is a diagonal matrix
@AkivaWeinberger I want to show $|\frac{f(x)-f(y)}{x-y}| \leq M$. That looks very close to being the derivative... if I fix x and let y -> x. The derivatives are all bounded. But I don't know how that helps me.
Let f be a continuous function that maps the unit interval [0, 1] in R to itself. Assume that f has a derivative f' which is defined and continuous on [0, 1] and that |f'(x)| < 1 for x ∈ [0, 1].
Show that there is a constant M < 1 such that for all x, y in [0, 1], |f(x) − f(y)| ≤ M|x − y|
@LeakyNun it's an $\Bbb R$-algebra homomorphism if we restrict the domain to a certain subalgebra of $\Bbb R(X)$ (the localization at kernel of $\Bbb R[x]\to\Bbb R$)
@MatheinBoulomenos @anon it's kind of paradoxical to me: if it isn't an eigenvector, then how do the set of eigenvectors (with the same eigenvalue) form a subspace?
@Daminark I think a theory is good iff it is complete, so examples are torsion-free divisible abelian groups, presburger arithmetic, dense linear-ordering without endpoints, and algebraically-closed field of fixed characteristic
