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
    Jan 23, 2017 at 15:24
  • 3
    $\begingroup$ 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. $\endgroup$ 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? $\endgroup$ 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. $\endgroup$ 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 $$ T^\hdual = \mathrm E_X^{-1} T^* \mathrm E_Y^{\vphantom{-1}} $$

  • 1
    $\begingroup$ 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
    Jan 24, 2017 at 7:31
  • $\begingroup$ how can we show this $\endgroup$ 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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.