Are super-operators always writable in a basis independent form?

[spocchio]

in particular, i wonder if the trasposition super operator is basis independent or not.

We can always write an operator W as
$\hat{W}=\sum_{i,j} c_{i,j} |i\rangle\langle j|$
and for the transposed we obtain
$\hat{W}^T=\sum_{i,j} c_{j,i} |i\rangle\langle j|$
we obtain a relation true for each \psi,\phi
$\langle\psi|\hat{W}^T|\phi\rangle=\langle\phi|\hat{W}|\psi\rangle$ (1)
Now let's write the transposition super-operator as
$\Lambda[\hat{W}]=\sum_{i,j} |i\rangle\langle j| \hat{W} |i\rangle\langle j| = \hat{W}^T$

Now seems that the transposed matrix does not depend from the basis,
I have to admit it because the formula (1) is true for each vector and don't depend from the basis.
But that's sound strange to me, because if I have an operator, it can always be diagonal in a certain basis, and so it is equal to its transposition.
We can always write W diagonal in a certain basis $\{|\tilde{k}>\}$ not neccesary equal to $\{|i>\}$:
$\hat{W}=\sum_{k} c_k |\tilde{k}\rangle\langle\tilde{k}|$
and from here do i get
$\hat{W}^T=(\sum_{k} c_k |\tilde{k}\rangle\langle\tilde{k}|)^T=(\sum_{k} c_k |\tilde{k}\rangle\langle\tilde{k}|=\hat{W}$

Where I'm wrong?
Can I write every super operator in a representation that is independent from the basis?

 

[azntoon]

Yes, you can write every super operator in a representation that is independent from the basis. The key is to understand that the transpose of an operator is always basis independent. This means that if you have an operator written in a particular basis, you can always transpose it without changing the outcome. This is because the transpose operation does not depend on the basis: it simply switches the positions of the matrix elements. As a result, the transposed operator remains the same regardless of what basis it is written in.