# Matrix representation of the adjoint of an operator, the same as the complex conjugate of the transpose of that operator?

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

• This is only true if you represent the operator with respect to an orthonormal basis I think. – Giuseppe Negro May 30 '13 at 8:43
• There is no specification of that in the question, I guess it is assumed, most of the questions in the textbook are, I believe, with respect to an orthonormal basis, maybe that'll change further in the book – aNxello May 30 '13 at 8:46

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.
in a complex vector space $V$ endowed with an inner product $\langle\,|\,\rangle$, the adjoint $\hat{A}^\star$ of the (linear) operator $\hat{A}$ is the unique operator with the property that $$\langle \hat{A}\psi | \varphi\rangle = \langle \psi | \hat{A}^\star\varphi\rangle,\qquad \forall \psi, \varphi\in V.$$
If that is the case, then a way to prove your claim easily is to fix an orthonormal base $(e_i\ :\ i=1\ldots N)$ and observe than any operator is uniquely determined as a complex linear combination of the operators $\hat{A}_{ij}$, defined by this property: $$\hat{A}_{ij}e_i=e_j\quad \text{and}\quad \hat{A}_{ij}e_k=0\quad \forall k\ne i.$$ This operator corresponds to a matrix of all zeros except for a $1$ at $(i, j)$ place. Now since the operation $\hat{A}\to \hat{A}^\star$ is (conjugate-)linear, you only have to check that the adjoint $\hat{A}^\star_{ij}$ is represented by the transpose of this matrix, and this is a matter of applying the definition.