real spectrum clarification

Question

Let $T: X\to X$ be a bounded operator on the real Banach space $X$.Does the spectrum of $T$ consist of the reals in the spectrum of its complexification?Or are they the same thing by definition?

Answer 1

If $T$ is invertible, then so is $T'$: indeed, if $ST=TS=I$, then $$ S'T'(x+iy)=STx+iSTy=x+iy. $$ Conversely, if $KT'=T'K=I$, then define an operator $S$ on $X$ by $Sx=K(x+i0)$. Then $$ STx=K(Tx+iT0)=KT'(x+i0)=x+i0. $$ In both cases the induced inverse is bounded by the Open Mapping Theorem.

It follows that if $\lambda\in\mathbb R$, then $T-\lambda I$ is invertible if and only if $T'-\lambda I$ is invertible; so $$ \sigma(T')\cap \mathbb R=\sigma(T). $$