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## 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.