Why does complex conjugation permute the rows (columns) of a character table

Question

If $\chi$ is the character of $\rho$, then $\overline{\chi}$ is the character of $\rho^*$ (dual) and $\chi_{irreducible} \iff \overline{\chi_{irreducible}}$.

This implies complex conjugation permutes the rows of a character table.

I cannot see why it is implied that complex conjugation would permute the rows of a character table. Why should the dual also be in the character table?

Nor can I see why complex conjugation swaps columns. It is just stated in my notes without any explanation.

I cannot find a good explanation in notes online

Answer 1

To see that the "dual" of an irreducible character is indeed an irreducible character:

Hint (mechanical).
Think about row orthogonality and how the inner product is defined (remember the set of irreducible characters is an orthonormal basis for the set of class functions).

Hint (representations).
Define a map $GL(V) \to GL(V)$ by sending a matrix $(a_{ij})$ to $(\overline{a_{ij}})$, that is, taking the complex conjugate of each of its entries. Is this an automorphism of $GL(V)$ when $V$ is a vector space over $\Bbb C$?

Answer 2

I'm assuming we are talking about finite groups. Then we have: $\rho(g)$ is orthogonal, therefore diagonalizable. Since $\rho(g)^{|G|} = 1$ the eigenvalues of $\rho(g)$ are roots of 1, this means that their inverses, which are the eigenvalues of $\rho(g^{-1})$ are their conjugates and therefore $\overline{\chi(g)}$ is $\chi(g^{-1})$