I have read that there are two 'options' for an adjoint when dealing with Hilbert spaces. Let $T : X \to Y$ be a bounded linear operator between the Hilbert spaces $X$ and $Y$.

  1. The Hilbert space adjoint: Define $T^* : Y \to X$ by $( T^*(y), x)_X = (y, T x)_Y$.

  2. The "usual" Banach space adjoint: Define $T^* : Y' \to X'$ by $\langle T^*(y^*), x \rangle_{X',X} = \langle y^*, T x\rangle_{Y',Y}$.

Here, $(\cdot,\cdot)_X$ refers to the scalar product in $X$, whereas $\langle \cdot, \cdot \rangle_{X',X}$ refers to the duality product between $X'$ and $X$.

It seems to me that these are both exactly the same, its just that we identify the duality product with the scalar product because we are in a Hilbert space and have an inner product at our disposal. Is this correct or are the adjoints in 1. and 2. fundamentally different?

  • 1
    $\begingroup$ You are right: applying the usual adjoint to a Hilbrrt space operator and identifying $ Y $ with $ Y'$ and $ X $ with $ X'$ (through the Riesz isometry) yields the Hilbert space adjoint $\endgroup$
    – Bananach
    Commented Jan 23, 2017 at 15:24
  • 4
    $\begingroup$ You need to be careful since the bijection between $X$ and $X^*$ is anti-linear. For example when you want to understand the finite-rank operators on a Hilbert space as the tensor product between the space and its dual then this problem arises. $\endgroup$ Commented Jan 23, 2017 at 19:02
  • $\begingroup$ @SebastianBechtel So when people say 'we identify $X$ with its dual $X^\ast$' that identification is not a regular isomorphsim? How does it come about that the bijection has to anti-linear anyway? $\endgroup$ Commented Jan 25, 2017 at 8:42
  • $\begingroup$ @eurocoder at least in the case of complex Hilbert spaces it is only an isometric anti-linear bijection. That it is anti-linear is due to the anti-linearity of the inner product in one component. $\endgroup$ Commented Jan 25, 2017 at 13:09

2 Answers 2


$\newcommand{\hdual}{\mathsf H}$Let $T^\hdual$ denote the "Hilbertian" adjoint (usually called conjugate transpose) and $T^*$ the "Dual" or "Banachian" adjoint. Let $\mathrm E_X:X \to X^*$ be the natural embedding of a Hilbert space onto its dual, namely $x \mapsto \langle x,\cdot \rangle$. Here the inner product is taken to be linear in the second component and conjugate-linear in the first.

We can show that $$ T^\hdual = \mathrm E_X^{-1} T^* \mathrm E_Y^{\vphantom{-1}} $$

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    $\begingroup$ One should $(\lambda T)^*=\lambda T^*$ and $(\lambda T)^H = \bar \lambda T^H$, which is a consequence of conjugate-linearity of $E_X$. $\endgroup$
    – daw
    Commented Jan 24, 2017 at 7:31
  • $\begingroup$ how can we show this $\endgroup$ Commented Sep 2, 2020 at 5:10

They are not fundamentally different.

But the difference exists and comes into focus when one writes down the formula $T^*T$. This composition makes sense for the Hilbert space adjoint, but not for the Banach space adjoint, since the domain of the latter is $Y'$ and not $Y$.

It's true, as Bananach said, that the Hilbert space adjoint is the composition of the Banach space adjoint with two anti-linear duality maps (so it ends up being linear).


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