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I've read an article which features the transpose of a bounded linear operator $A$, denoted by $A^{t}$. I have no idea what is this $A^{t}$ all about. Is this actually an adjoint? Are the transpose of an operator and the adjoint of an operator the same thing? Can you please help me on this. I'm a bit confused. Kindly look at Theorems 3.3 and 3.4 of this article. http://www.scielo.cl/pdf/proy/v33n4/art04.pdf

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In a finite dimensional vector space the adjoint $A^\dagger$ of a linear operator $A$ is represented by the complex conjugate of the transpose matrix that represents $A$ ( and this is true in any basis). This implies that in a real vector space theadjoint is simply the transpose.

For an infinite dimensional space the ''transpose'' of an operatore cannot be defined in terms of representative matrix, so we define the adjoint of a linear operator $S$ as the vector $A^\dagger$ such that $$ \langle A^\dagger x, y\rangle = \langle x,A y\rangle $$

where $\langle \cdot, \cdot \rangle $ is the inner product in the Hilbert space.

With an abuse of language, if the Hilbert space is over the reals, we can call this operator $A^\dagger$ the ''transpose'' of $A$.

As an example, if $A=\frac {d}{dx}$ on the space of real valued continuous functions on $[0,1]$ and such that $f(0)=f(1)=0$, with the inner product $\langle f(x),g(x)\rangle=\int_0^1 f(x)g(x)dx$, we can prove, using integration by part, that $A^\dagger =-\frac{d}{dx}$ and we can call this the ''transpose'' of $A$.

But note that the $A^\dagger$ depends not only from the operator, but also from the vector space and from the inner product.

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  • $\begingroup$ so what's gonna happen if the Hilbert space is over the field of complex numbers? the so-called "transpose" is no longer the adjoint? $\endgroup$ – Dong North Jun 17 '17 at 7:57
  • $\begingroup$ I think that in this case the use of ''transpose'' is misleading. $\endgroup$ – Emilio Novati Jun 17 '17 at 8:01
  • $\begingroup$ in an infinite dimensional Hilbert space (or any vector space in general), does a "transpose" still need an orthonormal basis of the space? That's what has been written on the article that i cited in my question, particularly on Theorems 3.3 and 3.4. $\endgroup$ – Dong North Jun 17 '17 at 8:09
  • $\begingroup$ I have read one definition about the transpose. It looks like the adjoint but not quite. The definition goes something like this. The transpose of $A$ is the operator $B$ such that $\langle Ax, y \rangle = \langle By, x \rangle$. This is for the field of complex scalars. But for the field of reals, the transpose becomes the adjoint since the conjugate of $\langle By, x \rangle$ is just $\langle x, By \rangle$. How do you find this definition? is this precise? $\endgroup$ – Dong North Jun 17 '17 at 8:26
  • $\begingroup$ Well,since the inner product is commutative, it seems the same definition as in my answer, with your $B$ as my $A^\dagger$. $\endgroup$ – Emilio Novati Jun 17 '17 at 12:03

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