Martin Argerami
Reputation
56,672
99/100 score
 Jan 14 revised Von Neumann algebras with uncountable sets of incompatible projections added 604 characters in body Jan 13 answered Von Neumann algebras with uncountable sets of incompatible projections Jan 13 answered To find Wronskian of a ODE Jan 13 awarded Yearling Jan 9 comment Proof that $A^{-1}=adj(A)/|A|$ I don't know the actual history, but things like this are usually a combination of the ideas of several very smart people working first on examples and abstracting from there. I Jan 9 answered Proof that $A^{-1}=adj(A)/|A|$ Jan 8 comment union of group von neumann To talk about "closure" you need an environment. Where would the union live? This is key to answer the question. Jan 8 comment solve $-u''(x)+\int_0^{\pi}u(y)dy=\lambda u(x)$,where $u(0)=u(\pi)=0$ I fail to see why you introduced the initial condition $u'(0)=1$. Jan 7 comment The series of $\sin A$, when $A$ is a $n \times n$ matrix The first inequality is not true. It fails for example for $n=1$ and $A=3\pi/2$. Jan 6 comment If$p \in B(H)$ is a projection, then $r \in A'$ if and only if the closed vector subspace $p(H)$ of $H$ is invariant for $A$. I feel you are making it unnecessarily complicated: from $pT^*p=T^*p$ you get $pTp=pT$ directly by taking adjoints. The assertion "W invariant under $T^*$ implies $W^\perp$ invariant under $T$" is basically what needs to be proven. Jan 5 comment Two Banach spaces, if and only if criterion for range of closed unbounded operator to be closed? Nice! $\ \ \ \$ Jan 5 comment Two Banach spaces, if and only if criterion for range of closed unbounded operator to be closed? Problems I see: 1. The distance to a closed subspace is not always realized; 2. If you have the inequality $\|x_n\|\leq\|y_n\|$ and $\{y_n\}$ is Cauchy, you cannot conclude that $\|x_n\|$ is Cauchy; 3. In your last part, your argument "works" for any closed operator: if $u_n$ is Cauchy in $\|\cdot\|_G$ then from the graph closed you would get your conclusion. Jan 5 answered Prove: the density operator of a pure state has exactly one non-zero eigenvalue equal to unity Jan 5 answered nilradical of a finite-dimensional algebra Jan 2 comment If a set has infinite measure is it NOT lebesgue measurable? If so why? @CarlMummert: good point! Jan 2 answered Prove that $\sqrt{2} \notin \mathbb{Q}(\sqrt{3})$. Dec 31 comment What do instantaneous rates of change really represent? @Bernard: +1; but to be pedantic, it measures rpm on the other side of the gear box ;) Dec 31 awarded Good Answer Dec 31 answered Applying Neumann series Dec 30 answered Set of discontinuities of a function that has both limits at each point of $R$