Inner Prod Space, Positive Definite Proof

[TaylorWatts]

Homework Statement

Let H be an inner product space.

Let T->H be a linear, self adjoint, positive definite operator.

Fix h in H and let g = T(h) / square root (1 + (T(h),h)). for h in H

Show that the operator S->H defined by S(v) = T(v) - (v,g)g for v in H is positive definite (and self adjoint, but I already proved that part).

Homework Equations

L is self adjoint if L = L*
and by proposition FOR ANY L, (L(v),u) = (v,L*(u))
but => (L(v),u) = (v,L(u)) for self adjoint operators.

L is a positive definite operator <=> (L(v),v) > 0 for all v in V (where V is an inner product space).

The Attempt at a Solution

So far I've tried to show (S(v),v) > 0 by substituting in T(v) - (v,g)g.

When I do that I get:

(S(v),v) = (T(v) - (v,g)g,v) =

(T(v),v) - ((v,g)g,v)= (T(v),v) - (v,g)*(g,v).

Then I substitute in for g:

(T(v),v) - (v,T(h)/square root(1 + (T(h),h))*(T(h)/square root(1 + T(h),h)),v)

And I realize (since T is positive definite) that the above expression is greater than:

(T(v),v) - (v,T(h))*(T(h),v).
since square root 1 + a positive number > 1, and pulling the 1/square root(1+(T(h),h)) out of the two inner products.

I don't see this as helping all that much though and I am now stuck.

 

[mighty2000]

Thank you for your question. Let me walk you through the proof step by step to help you understand it better.

First, we want to show that S is self adjoint. This means we need to show that for any two vectors v and u in H, (S(v),u) = (v,S(u)).

So let's start with (S(v),u):

(S(v),u) = (T(v) - (v,g)g, u)
= (T(v),u) - ((v,g)g,u)
= (T(v),u) - (v,g)*(g,u) (using the property you mentioned in your attempt at a solution)
= (T(v),u) - (v,g)*((T(h)/sqrt(1+(T(h),h)),u) (substituting in for g)
= (T(v),u) - (v,T(h))*(T(h),u)/sqrt(1+(T(h),h)) (since g = T(h)/sqrt(1+(T(h),h)))

Next, let's look at (v,S(u)):

(v,S(u)) = (v, T(u) - (u,g)g)
= (v,T(u)) - (v,(u,g))g
= (v,T(u)) - (u,g)*(v,g) (using the property again)
= (T(u),v) - (u,g)*(g,v) (since T is self adjoint)
= (T(u),v) - (u,T(h))*(T(h),v)/sqrt(1+(T(h),h)) (substituting in for g)

Now, we can see that (S(v),u) = (v,S(u)) if (T(v),u) = (T(u),v). This is true because T is self adjoint. Therefore, S is self adjoint.

Next, we want to show that S is positive definite. This means we need to show that (S(v),v) > 0 for all v in H.

(S(v),v) = (T(v),v) - (v,g)*(g,v)
= (T(v),v) - (v,T(h))*(T(h),v)/sqrt(1+(T(h),h)) (since g = T(h)/sqrt(1+(T(h),