Since I'm not taking summer classes I decided to do some self learning on more advanced mathematics, and I've found myself stuck on this problem:
I have to show that for any operator $\hat{A}$ the matrix representation of the adjoint $\hat{A}\dagger$ is given as the complex conjugate of the transpose of the matrix representation of the operator $\hat{A}$.
So basically show that
$\underline{A}\dagger = (\underline{A^T})^*$
I've done many examples on the textbook using actual matrices, but I don't really know how to prove this in terms of operators, any hint or where to start, or some guidance?
Thank you
Best Answer
Let $V,W$ be complex, finite dimenstional Hilbert spaces, $A \colon V \to W$ a linear operator. Let $(e_1, \ldots, e_n)$ and $(f_1, \ldots, f_m)$ be orthonormal bases of $V$ and $W$ respectively. If then $(\alpha_{ij})$ is the matrix representation of $A$ with respect to these bases, we have $$ Ae_i = \sum_j \alpha_{ij}f_j, \quad 1 \le i \le n $$ Or, as $(f_j)$ is orthonormal $$ (Ae_i, f_j) = \alpha_{ij}, \quad 1 \le i \le n, 1 \le j \le m $$ Let $(\beta_{ij})$ be the matrix representation of $A^\dagger$, we have then, as above $$ A^\dagger f_i = \sum_i \beta_{ij}e_j $$ or $$ (A^\dagger f_i, e_j) = \beta_{ij}, \text{ each $i,j$} $$ We get, by definition of the adjoint, that \begin{align*} \beta_{ij} &= (A^\dagger f_i, e_j)\\ &= (f_i, Ae_j)\\ &= \overline{(Ae_j, f_i)}\\ &= \overline{\alpha_{ji}} \end{align*} So $(\beta_{ij})_{i,j} = (\bar \alpha_{ji})_{i,j}$, as wanted.