Let $M_n$ denote the space of $n\times n$ matrices over complex numbers. The space of self-adjoint matrices is denoted $$ M_n^{sa} = \{A\in M_n\, :\, A^*=A \}, $$ where $A^*$ denotes the conjugate transpose of $A$, and the cone of positive semidefinite matrices will be denoted $$ M_n^+ = \{A\in M_n^{sa}\, :\, A\geq0\}, $$ where $A\geq0$ denotes that $A$ is positive semidefinite. The adjoint of a linear map of matrices $\phi:M_n\rightarrow M_m$ is the unique map $\phi^*:M_m\rightarrow M_n$ that satisfies $$ \mathrm{Tr}(\phi(A)B) = \mathrm{Tr}(A\phi^*(B)) $$ for all $A\in M_n$ and $B\in M_m$. The dual cone of a subset $S\subseteq M_{n}^{sa}$ is defined by $$ S^* = \{B\in M_{n}^{sa}\, : \, \forall A\in S, \, \mathrm{Tr}(AB)\geq 0\} $$ Note that $S^*$ is always a closed and convex cone.

Let $\phi$ be a linear map of matrices and consider the cones of self-adjoint matrices defined by $$ K = \{A\in M_n^{sa} \, :\, \phi(A)\geq 0\} $$ and $$ L= \{\phi^*(B) \, :\, B\geq 0\}. $$ It is evident that both $K$ and $L$ are both convex cones and that $L\subseteq K^*$ and $K\subseteq L^*$.

Question: Is it necessarily the case that $K^*=L$ or $L^*=K$? If not, under what circumstances would it hold that $K^*=L$ or $L^*=K$? For the linear maps $\phi$ that I've played around with so far this seems to be the case and I'm wondering if there is a counterexample.


We have $K = \phi^{-1}(M_n^+)$ and $L = \phi^*(M_n^+),$ where $\phi^{-1}(S)$ is notation for $\{x : \phi(x) \in S\}$ whether $\phi$ is invertible or not. So the question is whether or not $\phi^*(M_n^+)^* = \phi^{-1}(M_n^+)$ and $\phi^{-1}(M_n^+)^* = \phi^*(M_n^+)$

It's easier to keep track of what to expect when working in more generality, so consider a linear map $\phi:V \to W$, where $V$ and $W$ are finite-dimensional inner product spaces over $\mathbb R$. If you're comfortable with dual spaces, you could work in even more generality and drop the inner product structure.

Given a subset $S \subseteq W$, it's not sensible to expect a nice relationship in general between $\phi^{-1}(S)^*$ and $\phi^*(S)$. For example, adding elements to $S$ makes one of those sets smaller but the other one larger.

But it is sensible to expect one between $\phi^{-1}(S^*)$ and $\phi^*(S)^*$. Indeed,

$$ \phi^*(S)^* = \{v \in V : \langle v, \phi^*(s)\rangle \geq 0\;\forall s\in S \} = \{v \in V : \langle \phi(v), s\rangle \geq 0\;\forall s\in S \} = \phi^{-1}(S^*). $$

In your situation, we have $V = W = M_n^{sa}$ with inner product $\langle A, B \rangle = \operatorname{Tr}(AB)$, and $S = M_n^+$. Note that ${M_n^+}^* = M_n^+$. So we have

$$L^* = \phi^*(M_n^+)^* = \phi^{-1}({M_n^+}^*) = \phi^{-1}(M_n^+) = K.$$

From this point of view, the missing ingredient is the observation that ${M_n^+}^* = M_n^+$, not just ${M_n^+}^* \supseteq M_n^+$.

It follows that $K^* = L^{**} = \bar L,$ so $K^* = L$ if and only if $L$ is closed, which isn't always the case, e.g. the counterexample in the comments.

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    $\begingroup$ Ah, thanks! However, while $K$ is necessarily closed, it is not necessarily the case that $L$ is closed. To correct your answer, we always have that $L^*=K$ but $K^*=L^{**}=\overline{L}$, where $\overline{L}$ denotes the closure of $L$. $\endgroup$ – Luftbahnfahrer Nov 7 '17 at 17:08
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    $\begingroup$ As an aside to show that $L$ is not necessarily closed, consider the map $\phi:M_2\rightarrow M_2$ defined by $ \phi\left(\begin{array}{cc} w & x\\ y & z\end{array}\right) \mapsto \left(\begin{array}{cc} w & z\\ z & 0\end{array}\right) $. $\endgroup$ – Luftbahnfahrer Nov 7 '17 at 17:10

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