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Can someone tell me if this proof is correct?

Claim:V is a vector space over the Complex field. $T:V\rightarrow V$ is a normal operator. Then if $v\in V$ is an eigenvector with the eigenvalue $\lambda \in C$ then , v is eigenvector of $T^*$ and $\overline{\lambda}$ is its eigenvalue.

Proof: $\langle Tv,v\rangle=\langle \lambda v,v\rangle=\overline{\lambda}\langle v,v\rangle=\langle v,\overline{\lambda}v\rangle$

On the other hand

$\langle Tv,v\rangle=\langle v,T^*v\rangle$ and therefore we have $\langle v,T^*v\rangle=\langle v,\overline{\lambda}v\rangle$ for every eigenvector v that belongs to the eigenvalue $\lambda$. We can notate $V_{\lambda}$ as the space of eigenvectors of $\lambda$.

If we reduce the domain from V to $V_{\lambda}$ we will get that

$\langle v,T_{|V_{\lambda}}^{*}v\rangle=\langle v,\overline{\lambda}v\rangle$ is true for every $v \in V$ and then we will get

$\langle v,(T_{|V_{\lambda}}^{*}-\overline\lambda Id)v\rangle=0$ for every $v\in V_{\lambda}$ and therefore, $T_{|V_{\lambda}}^{*}=\overline\lambda Id$ for every $v\in V_{\lambda}$ as desired.

Jonas Meyer
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Charles Carmichael
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    How do you know that $\langle v, (T^-\overline{\lambda}Id)v\rangle = 0$ for all $v$ implies that $T^-\overline{\lambda}Id = 0$? – Santiago Canez Jul 27 '14 at 18:01
  • I should have written that. We have proved that if V is a vector space over the field C (Complex field) and we have an operator such that $<Sv,v>=0$ $\forall v \in V$ then S=0. So I thought I could reduce my vector space to the eigenvector space with $\lambda$ eigenvalue and I can use that. But I am not sure I can actually do that.. – Charles Carmichael Jul 27 '14 at 18:10
  • @SantiagoCanez Could it possibly be using a consequence of the Hahn--Banach theorem? – Chris Cave Jul 27 '14 at 18:10

1 Answers1

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If $T$ is normal on a Complex space $X$, then $$ \|Tx\|^{2}=\|T^{\star}x\|^{2},\;\;\; x \in X. $$ Because $\lambda I$ is normal with adjoint $\overline{\lambda}I$, then $\|(T-\lambda I)x\|=\|(T^{\star}-\overline{\lambda}I)x\|$ also holds for all $x$ if $T$ is normal. That's the missing piece.

Disintegrating By Parts
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  • Now I noticed that I haven't used the fact that our operator is normal at any point... Why using this is essential in the proof? In other words, assuming T is not normal, where would have this failed? – Charles Carmichael Jul 27 '14 at 18:24
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    Yes, if $T$ is not normal, something else can happen. Suppose $\mathcal{N}(T-\lambda I)\ne {0}$ and also suppose that $\mathcal{R}(T-\lambda I)=X$. Then $\mathcal{N}(T^{\star}-\overline{\lambda} I)=\mathcal{R}(T-\lambda I)^{\perp}={0}$ even though $\mathcal{N}(T-\lambda I)\ne{0}$. The normal condition forces $\mathcal{N}(T-\lambda I)=\mathcal{N}(T^{\star}-\overline{\lambda}I)$, which is a very strong condition. – Disintegrating By Parts Jul 27 '14 at 18:37
  • Can you clarify what N and R is? Thanks for the feedback by the way it really helped! – Charles Carmichael Jul 27 '14 at 19:36
  • $\mathcal{N}$ means null space, and $\mathcal{R}$ means range. – Disintegrating By Parts Jul 27 '14 at 19:45
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    A concrete example: Let $X=\mathcal{l}^{2}$ be the linear space of infinite, square-summable sequences $(a_{0},a_{1},a_{2},\cdots)$. Define $T(a_{0},a_{1},a_{2},\cdots)=(a_{1},a_{2},a_{3},\cdots)$. Then $T(1,0,0,0,\cdots)=(0,0,0,\cdots)=0$, but $T^{\star}$ is the right shift $T^{\star}(a_{0},a_{1},a_{2},\cdots)=(0,a_{0},a_{1},a_{2},\cdots)$ which has no null space. So $0$ is an eigenvalue of $T$, but not of $T^{\star}$. This is an example where $TT^{\star}-T^{\star}T$ is not $0$, but is zero except for the one-dimensional linear space spanned by $(1,0,0,0,\cdots)$. – Disintegrating By Parts Jul 27 '14 at 19:49