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I am looking for a formal definition of the transpose of a linear operator (if there is). I've read an article about linear preservers and the author used the notation $A^{Tr}$ to denote the transpose of an operator $A$ (on Hilbert space $H$) with respect to a fixed but arbitrary orthonormal basis of $H$. I don't know if the transpose $A^{Tr}$ that he's talkin' about is actually the adjoint of an operator $A$, but he just simply used different terminology. I'm a bit confused. Can you please help me on this. Does an operator have a transpose? or it is just the same as the adjoint?

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The 'obvious' definition (generalizing from the finite dimensional case) is that given the orthonormal basis $v_1, v_2, \ldots$ we define $A^{Tr}$ as the unique operator $B$ such that $\langle B v_i, v_j \rangle = \langle A v_j, v_i \rangle$ for all $i, j$.

The relation between this thing and the adjoint $A^*$ of $A$ is that $A^*$ can be seen as the complex conjugate of $A^{tr}$ and vice versa. This follows from the fact that $A^*$ is the unique operator $C$ such that $\langle A v_i, v_j \rangle = \langle v_i, Cv_j \rangle$ for all $i, j$, together with the fact that $\langle v, w \rangle = \overline{\langle w, v \rangle}$ for all $v, w \in H$.

In practice this means you have two ways of computing $A^{tr}v$ for some $v \in H$: either you use the above formula, or you compute $A^*v$, write it out w.r.t. your special basis and then take the complex conjugate of each coefficient.

Now the real mystery is why someone would introduce this notion of transpose. The adjoint is very natural because it can be defined purely in terms of the inner product, without choosing a basis. It also has a nice generalization to maps between different vector spaces that need not even be Hilbert spaces. This transpose on the other hand is a little weird because it depends really on the particular basis of your very Hilberty Hilbert space. But perhaps the author has a good reason for looking at this?

Of course this all changes when the Hilbert space is defined over the real numbers instead of the complex numbers. Then complex conjugation does nothing and hence the transpose and the adjoint coincide.

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  • $\begingroup$ Please have a look at Theorem 3.3, 3.4 and 4.2 of this article. ac.els-cdn.com/S131951661400005X/… This is the one that lead me into confusion. I really have no idea what $A^{Tr}$ pertains to. I appreciate it if you could help me on this. I am doing a research now in linear preservers and this thing about $A^{Tr}$ bothers me.. $\endgroup$ – Dong North Jun 15 '17 at 9:40
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For an $n\times n$ matrix $A$, the adjoint $A^*$ is just the complex conjugate of the transpose (so, if the matrix has real entries the adjoint and the transpose are the same).

We cannot extend this definition to operator on an infinite dimensional Hilbert space, but we can define the adjoint $A^{\dagger}$ of an operator $A$ as the operator such that $(A^\dagger u,v)=(u,Av)$. Where $(\cdot,\cdot)$ is the inner product.

If the vector space is over $\mathbb{R}$ , so that the inner product is real valued, than we can say that $A^\dagger$ is the ''transpose'' of $A$. Note that, in general, given a linear operator on a space of functions the adjoint $A^\dagger$ depends not only from the operator, but also from the vector space in which it operate and from the inner product that is used.

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  • $\begingroup$ Please have a look at Theorems 3.3, 3.4 and 4.2 of this article. ac.els-cdn.com/S131951661400005X/… This is the one that lead me into confusion. I really have no idea what does this $A^{Tr}$ pertain to. I appreciate it if you could help me on this. I am doing a research now about linear preservers and this thing about $A^{Tr}$ bothers me.. $\endgroup$ – Dong North Jun 15 '17 at 9:42

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