Transpose of an operator with respect to a fixed but arbitrary orthonormal basis

Question

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

Answer 1

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.