6
$\begingroup$

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.

$\endgroup$
2
+50
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.