How do we show that $A^{\dagger^{\dagger}} = A $ without assuming $A$ to be a explicit matrix. That is, given a linear operator $A$, let us define $A^\dagger$ to be a unique operator such that $\langle A^\dagger u \mid v \rangle = \langle u\mid A v \rangle $ (standard inner product in Hilbert space) for any vectors $u$ and $v$. How can we prove that using this definition alone.

  • $\begingroup$ Can someone explain "$\dagger$" to me? $\endgroup$ – A. Chu Nov 3 '12 at 9:20
  • $\begingroup$ @jasoncube, see conjugate transpose and Hermitian adjoint. $\endgroup$ – Rahul Nov 5 '12 at 5:03
  • $\begingroup$ $A^*$ is a good notation to represent the adjoint of $A$. Because, in linear algebra $A^\dagger$ is used for moore-penrose inverse generally. $\endgroup$ – David Nov 25 '15 at 5:25

Per the definition, $A^{\dagger\dagger}$ is the unique linear operator such that $$\langle A^{\dagger\dagger}u\mid v\rangle = \langle u\mid A^\dagger v\rangle = \overline{\langle A^\dagger v\mid u\rangle} = \overline{\langle v\mid Au\rangle} = \overline{\overline{\langle Au\mid v\rangle}} = \langle Au\mid v\rangle$$ for all vectors $u$ and $v$, so...


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