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

  1. #1
    Member Last_Singularity's Avatar
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    T is a normal operator; prove that R(T) = R(T*)

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

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

    But what about $\displaystyle R(T)=R(T^*)$? I know that $\displaystyle R(T^*) = N(T)^{perp}$ but how does that help?
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  2. #2
    Senior Member Shanks's Avatar
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    $\displaystyle R(T^*) = N(T)^{perp}=N(T^*)^{perp}=R(T)$
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