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In my lecture notes, we have a linear map $\mathcal{A}(X)=X_{11}$ that maps $X\in S^2$ 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 $\mathcal{A}(X)= \begin{bmatrix}1&0\end{bmatrix} X \begin{bmatrix}1\\0\end{bmatrix} =X_{11}$ , so we solve for $X$: $$ \begin{align} X&=\begin{bmatrix}1\\0\end{bmatrix} \Bigg(\begin{bmatrix}1&0\end{bmatrix} X \begin{bmatrix}1\\0\end{bmatrix}\Bigg) \begin{bmatrix}1&0\end{bmatrix}\\ &= \begin{bmatrix}1\\0\end{bmatrix} X_{11} \begin{bmatrix}1&0\end{bmatrix}\\ &=\mathcal{A}^*(X_{11}). \end{align}$$

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: $$\mathcal{A}^*(X_{11})= \begin{bmatrix}1&0\end{bmatrix}^T X_{11} \begin{bmatrix}1\\0\end{bmatrix}^T = \begin{bmatrix}1\\0\end{bmatrix} X_{11} \begin{bmatrix}1&0\end{bmatrix}. $$

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.

Dan
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  • The linear map is defined on vector space. In particular, if we choose orthonormal bases of both spaces, one can show that the matrix representations of $L^∗$ and $L$ are related by $[L^∗]=[L]^T$. You can read more about it here https://math.stackexchange.com/questions/1769834/definition-of-adjoint-of-a-linear-map – Avanish Singh Nov 18 '18 at 20:01

1 Answers1

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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:

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

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

Dan
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