Show that $\ker(T)^\bot= ran(T)$ when $T$ is self-adjoint on a Hilbert space.

Question

Problem: Let $H$ be a Hilbert space. If $T\in B(H)$ is a self-adjoint operator, then $\ker(T)^\bot= ran(T)$.

Attempt: I have already shown the usual accompaniment to this problem, which is to show that in the same context, $ran(T)^\bot=\ker(T)$. The proof I used there followed this reasoning. For $z\in H$, consider,

$$z\in ran(T)^\bot\iff z\,\bot\, ran(T)$$

$$\iff\langle z,y\rangle=0,\,\forall y \in ran(T)$$

$$\iff\langle z,Tx\rangle=0,\,\forall x \in H$$

$$\iff\langle Tz,x\rangle=0,\,\forall x \in H$$

$$\iff Tz=0$$

$$\iff z\in\ker(T)$$

I quite like this proof in that it simultaneously shows the inclusion of the former set in the latter and vice versa. I would like to employ a similar method to proving that $\ker(T)^\bot= ran(T)$, but I am running into difficulty. Like before, for $z\in H$,

$$z\in\ker(T)^\bot\iff z\bot\ker(T)$$

$$\iff\langle z,y\rangle=0,\,\forall y\in\ker(T)$$

Now $y\in\ker(T)$ means that $Ty=0$. How can I use this fact at this point in the proof or make more progress and introduce the operator, $T$?

Answer 1

Since $\ker(T) = {\rm ran}(T)^{\perp}$, we have $$\ker(T)^{\perp} = {\rm ran}(T)^{\perp\perp} = \overline{{\rm ran}(T)}.$$ So the result is true if and only if ${\rm ran}(T)$ is closed.