# Self Adjoint operators and Normal operators.

I have this problem I am working on:

$H$ is an inner product space with inner product $(\cdot , \cdot)$ over a subfield of the complex numbers.

Suppose $T\in L(H,H)$ has an adjoint $T^*$. Let R=$T+T^*$ and S=$T-T^*$. I'm supposed to show that R and S are normal operators and that $T ∘T^*$ and $T^*∘T$ are self adjoint.

So for the first part, to show they are normal operators, this is what I have:

For $R$: Need to show $(R(\alpha),\beta)=(\alpha,R(\beta))$ $$\implies ((T+T^*)(\alpha),\beta)=(T(\alpha)+(T^*(\alpha),(\beta))$$

$$=(T(\alpha),\beta)+(T^*(\alpha),\beta)$$

$$=(\alpha,T^*(\beta))+(\alpha,T(\beta))$$

$$=(\alpha,(T^*+T)(\beta))$$

$$=(\alpha,R(\beta))$$

I did something similar for $S$. Did I show that they are normal operators or that they are self adjoint? How do I show that $T \circ T^*$ and $T^*\circ T$ are self adjoint? Thanks!

• The identity $(R\alpha, \beta)=(\alpha, R\beta)$---equivalently, $R=R^*$---defines self-adjointness, not normality. Normality means $R\circ R^*=R^*\circ R$. Clearly, anything self-adjoint is normal. – Peter Franek Nov 7 '14 at 18:02
• So since I showed that R and S are self adjoint, would it be sufficient to say that since any self-adjoint is normal, then R and S are normal? – user141745 Nov 7 '14 at 19:18
• $S$ is not self-adjoint – daw Nov 7 '14 at 19:19
• For S I have that I need to show: $(S(\alpha),\beta)=(\alpha,S(\beta)$ -->$((T-T^*)(\alpha),\beta)$=$(T(\alpha)-T^*(\alpha),(\beta))$ =$(T(\alpha),\beta)-(T^*(\alpha),\beta))$ =$(\alpha,T^*(\beta))-(\alpha,T(\beta))$ =$(\alpha,T^*(\beta))-(\alpha,T(\beta))$ =$(\alpha,T^*(\beta)-T(\beta))=(\alpha,(T^*-T)(\beta))$ =$(\alpha,S(\beta))$ Doesn't this show that S is self-adjoint? – user141745 Nov 7 '14 at 19:37
• No, it shows that $S^*=-S$. – Peter Franek Nov 7 '14 at 22:27