- #1
TaylorWatts
- 16
- 0
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.