Second Hermitian scalar product

[Ioiô]

Homework Statement

Let (u,v)1 be a second Hermitian scalar product on a vector space V.

Claim: There exists a positive transformation T with respect to the given scalar product (u,v) such that (u,v)1 = (Tu,v) for all u,v in V.

Homework Equations

A transformation T is positive if

1) T is self-adjoint
2) (Tv, v) greater than or equal to zero (and real, of course)

The Attempt at a Solution

I am confused at what we are trying to prove. Is T positive with respect to (u,v) or are we using the definition of (u,v)1 and showing it is positive?

Initially, my plan to prove the existence of T was to first choose a matrix with respect to some orthonormal basis and then show that T is positive with respect to the inner product (u,v)1. However, as mentioned, the claim is that T is positive with respect to (u,v). How does (u,v)1 fit into the problem? (What is meant by second "Hermitian product?")

Thanks

 

[mighty2000]

for any help!



Thank you for your post. I understand your confusion about the problem. Let me try to clarify things for you.

Firstly, a second Hermitian scalar product on a vector space V is a different Hermitian scalar product than the one that is given in the problem (u,v). This means that there are two scalar products defined on the same vector space V, (u,v) and (u,v)1.

The claim is that there exists a positive transformation T with respect to the given scalar product (u,v), which means that T satisfies the two conditions you mentioned: 1) T is self-adjoint and 2) (Tv,v) is greater than or equal to zero for all v in V.

However, this positive transformation T also has the property that (u,v)1 = (Tu,v) for all u,v in V. In other words, T is also positive with respect to the second Hermitian scalar product (u,v)1.

To prove this claim, you can use the definition of a positive transformation and show that T satisfies both conditions for both scalar products (u,v) and (u,v)1. This will prove the existence of such a transformation T.

I hope this helps. Let me know if you have any further questions.