# Bounded self-adjoint linear operator is injective…

Let $$T : H → H$$ be a bounded self–adjoint linear operator on a Hilbert space $$H$$. Suppose the range $$R(T)$$ is dense in $$H$$. Prove that $$T$$ is injective.

• I want to show that the null space only contains zero, but I am confused in where to go. – AverageMean Dec 5 '19 at 20:01

## 1 Answer

We have $$R(T)^{\perp}=\overline{R(H)}^{\perp}=H^{\perp}=\{0\}$$, but $$\ker T^{\ast}=(R(T))^{\perp}$$.

• So how do we know that the range is closed? We only know that the range of T is dense. – Overachiever Dec 12 '19 at 18:25
• Okay, just made a mistake, we don't need that. Just keep in mind that $A^{\perp}=\overline{A}^{\perp}$. – user284331 Dec 12 '19 at 18:30