Krdytkyu

In my lecture notes, we have a linear map $mathcal{A}(X)=X_{11}$ that maps $Xin S^2$ to its first element. It then claims that $mathcal{A}^*(X_{11})=
begin{bmatrix}
X_{11}&0\
0&0
end{bmatrix}$ is the adjoint.

To me I understand it by $mathcal{A}(X)=
begin{bmatrix}1&0end{bmatrix}
X
begin{bmatrix}1\0end{bmatrix}
=X_{11}$
, so we solve for $X$:
$$
begin{align}
X&=begin{bmatrix}1\0end{bmatrix}
Bigg(begin{bmatrix}1&0end{bmatrix}
X
begin{bmatrix}1\0end{bmatrix}Bigg)
begin{bmatrix}1&0end{bmatrix}\
&=
begin{bmatrix}1\0end{bmatrix}
X_{11}
begin{bmatrix}1&0end{bmatrix}\
&=mathcal{A}^*(X_{11}).
end{align}$$

However, (coming from a physics background) this seems like we found the inverse $mathcal{A}^{-1}$, not $mathcal{A}^*$ the adjoint. From what I know (from physics), the adjoint of an operator is its (conjugate) transpose, which we would have derived by doing:
$$mathcal{A}^*(X_{11})=
begin{bmatrix}1&0end{bmatrix}^T
X_{11}
begin{bmatrix}1\0end{bmatrix}^T
=
begin{bmatrix}1\0end{bmatrix}
X_{11}
begin{bmatrix}1&0end{bmatrix}.
$$

Is either of these the correct way? I'm guessing it's probably the latter, and that here we happen to have $mathcal{A}^*=mathcal{A}^{-1}$, but I'm confused because I've never seen this done before, and I've also never seen a map/operator not represented just by a single matrix.

I figured out the correct way to derive the answer. It is important to note (and something I forgot to mention above) that we are working with a trace inner product. We derive $mathcal{A}^*$ as follows:

$$
DeclareMathOperator{Tr}{Tr}
langle y,mathcal{A}(X)rangle
=langle mathcal{A}^*(y),Xrangle
=yX_{11}
=Trbig(mathcal{A}^*(y)Xbig)
=sum_{i,j} big(mathcal{A}^*(y)big)_{ij}X_{ij}
$$
Thus for a general $Xin S^2$, we must have that $mathcal{A}^*(y)=begin{bmatrix}y&0\0&0end{bmatrix}.$

I'm not $100$% certain about the other two ways I proposed, but this is the way to derive $mathcal{A}^*$ in general.