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I have been struggling to analyse the spectrum of the adjoint of a bounded linear operator on a hilbert space.
Throughout the internet I have found vague references that $\sigma(T^*) = \sigma(T)$ but I fail to be able to prove that.
What I have succeeded in doing is to prove that $\rho(T^*) = \overline{\rho(T)}$ (where $\rho$ is the resolvent $\rho(T) = \mathbb{C} \backslash \sigma(T)$).
Thus imho $\sigma(T^*)$ should be equal to $\overline{\sigma(T)}$ and not the above...

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  • $\begingroup$ This isn't true for Hilbert spaces (see msteve's answer). It is however true for Banach spaces. $\endgroup$ Feb 7 '17 at 10:05
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    $\begingroup$ @enthdegree How can that be, considering that every Hilbert space is a Banach space? $\endgroup$
    – iolo
    Feb 7 '17 at 18:39
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    $\begingroup$ Because adjoint operators in (complex) Hilbert spaces are constructed using an inner product $(\cdot | \cdot):\mathfrak{H}\times \mathfrak{H}^\ast \to \mathbb{C}$, which is sesquilinear. On the other hand the composition-with-dual used to construct Banach space adjoint operators, $\langle \cdot,\cdot\rangle:\mathfrak{X}\times \mathfrak{X}^\ast \to \mathbb{F}$, is an outright bilinear form. $\endgroup$ Feb 8 '17 at 1:31
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    $\begingroup$ To expand on @enthdegree's comment, these two different notions of adjoint can be reconciled: the Hilbert space adjoint is the Banach space adjoint using the natural identification of the dual of a Hilbert space with its conjugate space. This identification arises via an antilinear map, and hence the introduction of the conjugate. $\endgroup$
    – J. Loreaux
    Feb 9 '17 at 18:08
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You're correct that $\sigma(T^*) = \overline{\sigma(T)}$, where the overline denotes complex conjugation (and not closure as a subset of $\mathbb{C}$ - recall that the resolvent set of a bounded operator is open, and hence the spectrum is closed). In order to conclude that $\sigma(T^*) = \sigma(T)$, you need to impose the stronger condition that the eigenvalues be real (by specifying that $T$ be self-adjoint, e.g.).

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  • $\begingroup$ If the hilbert space is real, then the equality holds, right? Or in the real case these things are not well defined? $\endgroup$
    – Filburt
    Jan 26 '18 at 1:59
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The critical point is in complex Hilbert space you use a sesquilinear inner product to make adjoints but in Banach space you use a bilinear form. So in the Banach situation the conjugates disappear.

To reduce confusion I'll replace the two senses of adjoint with $T^T$ for Banach adjoint and $T^H$ for Hilbert adjoint. We borrow the rest of our notation from Pedersen:

Adjoint $T^{H}$ for $T$ in (complex) Hilbert space $\mathfrak{H}$ is constructed through a sesqui-linear inner product $(\cdot|\cdot):\mathfrak{H}\times \mathfrak{H}^\ast→\mathbb{C}$ as follows: $$T^H:\mathfrak{H}^\ast\to \mathfrak{H}^\ast; \qquad (\cdot|T^H\phi)\triangleq (T \cdot |\phi) \in \mathfrak{H}^\ast $$

On the other hand, in a Banach space $\mathfrak{X}$, adjoint $T^T$ of $T$ is constructed using a bi-linear composition-with-dual operator $\langle \cdot,\cdot\rangle :\mathfrak{X}\times \mathfrak{X}^\ast\to \mathbb{F}$ in an identical looking way:

$$T^T:\mathfrak{X}^\ast\to \mathfrak{X}^\ast; \qquad \langle \cdot, T^T\phi\rangle\triangleq\langle T\cdot ,\phi\rangle \in \mathfrak{X}^\ast$$

Spectrum is: $$\sigma(T)=\left\{ \lambda : \lambda I - T \not\in \mathbf{GL} \right\}.$$

$\mathbf{GL}$ is the general linear group, 'operators with bounded inverse.'

With this notation, it is true that $\sigma(T)=\sigma(T^T)=\overline{\sigma(T^H)}$.

We see $\sigma(T)=\overline{\sigma(T^H)}$ by computing: \begin{align} \sigma(T^H)&=\{\lambda: \lambda I^H - T^H \not\in \mathbf{GL}(\mathfrak{H^\ast})\} \\ &= \{\overline\lambda: \phi\mapsto\phi((\lambda I - T)(\cdot)) \not\in\mathbf{GL}(\mathfrak{H}^\ast)\}\\ &= \{\overline\lambda: (\lambda I - T) \not\in\mathbf{GL}(\mathfrak{H})\} \tag{A} \\ &= \overline{\sigma(T)} \end{align} where $(\text{A})$ follows because, letting $S^H=\phi \mapsto \phi((\lambda I-T)(\cdot))$, if $(\lambda I -T)\in\mathbf{GL}(\mathfrak{H})$, then by computation $\phi\mapsto \phi((\lambda I-T)^{-1}(\cdot))\in \mathbf{GL}(\mathfrak{H}^\ast)$ is an inverse of $(S^H)^{-1}$ so $S^H \in\mathbf{GL}(\mathfrak{H}^\ast)$. On the other hand $S^H\in\mathbf{GL}(\mathfrak{H}^\ast)\Rightarrow (S^H)^{-1}\in \mathbf{GL}(\mathfrak{H}^\ast)$ and by computation, $((S^H)^{-1})^H\in \mathbf{B}(\mathfrak{H})$ must behave as an inverse for $\lambda I - T.$ (We have just shown the complements of the two sets are subsets of one another)

The Banach part follows by the same argument, replacing $(\cdot|\cdot)$ with $\langle \cdot,\cdot\rangle$, $\mathfrak{H}$ with $\mathfrak{X}$, $\cdot^H$ with $\cdot ^T$, and getting rid of conjugation.

This is part of E 4.1.11 in Analysis Now by Pedersen.

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  • $\begingroup$ you make the notation more complicated, not simpler. $\endgroup$
    – Rüdiger
    Feb 9 '17 at 22:50
  • $\begingroup$ Unfortunately you need precisely all of it to make sense of the question $\endgroup$ Feb 10 '17 at 0:13
  • $\begingroup$ your explanation is so clear! thank you! $\endgroup$
    – allen i
    Apr 20 '20 at 18:26

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