Recent content by LPB

I see what you're saying. Could a simple counterexample for the complex case be A=[[0,i],[0,0]] and B=[[0,0],[i,0]] (keeping x=[a,b])? If the operator matrices are self-adjoint, it seems A would always equal B ... for instance, A=B=[[0,1],[1,0]] or A=B=[[0,i],[-i,0]]. Is this right? OK, so...

Homework Statement Let A and B be nxn positive self-adjoint matrices such that for all x \in Cn, x*Ax = x*Bx. Prove that A = B. Equivalently, prove that if A, B are positive operators on H such that <Ax,x> = <Bx,x> \forall x \in H, then A = B. Hint: See Lemma 2.12. Homework Equations...

Thank you - I think that makes sense! The explanation of (x*)Ax was especially helpful. I've rewritten the steps so I think they make sense to me. The basic "flow" that I ended up with is: (x*)Ax = <x,Ax> = <x,Tx> = <Tx,x> \geq 0 Does this seem right? Again, thank you - I really...

The definition given in my textbook for a positive operator is: Let T:H\rightarrowH be a linear operator. Then T is called positive if T is self-adjoing and <Tx,x> \geq 0 for all x \in H. We write T \geq 0 for any positive operator. I'm taking a course that suggests two semesters of linear...

Homework Statement Let T be a positive operator on a Hilbert space H. Pick an orthonormal basis {e1,e2,...,en} for H. Let A=[aij] be the nxn matrix representation of T with respect to the basis {e1,e2,...,en}, so that Tej=\sumaijei, j=1,2,...,n. (The summation is from i=1 to n; I'm not...