If $T$ Self Adjoint So T+iI is 1-1

Question

let assume that T+iI is not 1-1 so $-i$ is an eigenvalue, so $T$ is not self adjoint as self adjoint transformation has only real eigenvalues so is it true?

Answer 1

Here are the main facts:

• $T-\lambda I$ is not injective iff $\lambda$ is an eigenvalue of $T$.

• If $T$ is self-adjoint and $\lambda$ is an eigenvalue of $T$, then $\lambda$ is real.

Answer 2

If $v\neq0$ is such that $Tv=-iv$, then $0\neq\|v\|^2=i\langle Tv,v\rangle\in\mathbb{R}$.

This means that $-\langle Tv,v\rangle=\overline{\langle Tv,v\rangle}=\langle v,Tv\rangle$.

If, in addition, $\langle Tv,v\rangle=\langle v,Tv\rangle$, then $\langle Tv,v\rangle=0$, which is a contradiction.