T is a normal operator; prove that R(T) = R(T*)

**Last_Singularity** · November 24th 2009, 05:34 PM

Question: Let $T$ is a normal operator on a finite-dimensional inner product space $V$ . Show that $N(T)=N(T^*)$ and that $R(T)=R(T^*)$ .

I think that I got the first part. Suppose $x \in N(T)$ Then $T(x)=0$ . But $||T(x)|| = ||T^*(x)||$ for all $x$ so $T^*(x)=0$ as well. So $x \in N(T^*)$ , which means that $N(T) \subseteq N(T^*)$ . The same logic goes the other way around and we conclude $N(T) = N(T^*)$ .

But what about $R(T)=R(T^*)$ ? I know that $R(T^*) = N(T)^{perp}$ but how does that help?

**Shanks** · November 24th 2009, 07:39 PM

$R(T^*) = N(T)^{perp}=N(T^*)^{perp}=R(T)$

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T is a normal operator; prove that R(T) = R(T*)

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