1
$\begingroup$

Let $p,q:H\to H$ bounded, linear operator on a Hilbert space $H$, such that $p^*=p=p^2$ and $q^*=q=q^2$ ($p^*$ is the adjoint of $p$, for q the same. YOu call $p$ and $q$ a projection). Let $q(H)\subseteq p(H)$. Why is $\langle (p-q)(h),h\rangle \ge 0$ for all $h\in H$?

Here (bounded linear) orthogonal projections on Hilbert spaces is a discussion that $p$ is the identity on $im(p)$, zero on $im(p)^\perp$ and $q$ is the identity on $im(q)$, zero on $im(q)^\perp$. If $im(q)\subseteq im(p)$ and if you take $h\in H$ such that $h=h_1+h_2$ with $h_1\in im(p)$ and $h_2\in im(p)^\perp$, I have to do distinctions of cases now, right?

I think the proof shouldn't be difficult, but I'm stuck here. Can you explain me how to continue the proof?

$\endgroup$
1
$\begingroup$

Let $T = p-q$. Note that $p(H)$ and $p(H)^\perp$ are invariant subspaces of $T$. Consider the restriction of $T$ to $p(H)$. We can write $$ T\mid_{p(H)} = \operatorname{id} - q $$ It should be easy to show that this operator is positive (I'll leave it to you).

Of course, the restriction of $T$ to $p(H)^\perp$ is $0$, which is a positive operator.

Note that $p(H)$ is closed since $p(H) = \ker p - \operatorname{id}$, so that $H = p(H) \oplus p(H)^\perp$. Since $T$ is positive on invariant subspaces whose direct sum is $H$, $T$ is positive on $H$.

In fact, we could have proven that $T$ is additionally a projection onto $p(H) \cap q(H)^\perp$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.