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This is an open-ended question I have. Is there a function of two positive-semidefinite hermitian operators $\min(A,B)$ returning another positive-semidefinite Hermitian operator such that:

  1. If A and B commute, the eigenvectors of $\min(A,B)$ should be the shared eigenvectors of A and B, with eigenvalues $\min (\lambda_A , \lambda_B)$ where $\lambda_A$ , $\lambda_B$ are the corresponding eigenvalues for A and B.

  2. When A and B do not necessarily commute, $A - \min(A,B)$ and $B - \min(A,B)$ should always be positive semidefinite, so that both A and B are greater than $\min(A,B)$ in the Löwner order.

If yes, is there a straightforward proof or construction of such a function? If not, is there any theorem limiting what properties such a function can have?

Now, these two axioms by themselves are fairly weak. You can easily find a trivial but not so powerful example by defining $\min(A,B)$ as being defined by axiom 1) if the two operators exactly commute and $\min (\lambda^{\min}_A , \lambda^{\min}_B) I$ otherwise, where $\lambda^{\min}_A$, $ \lambda^{\min}_B$ are the smallest eigenvalues of A and B.

So I'll reformulate the question: is there a reasonable such function with more useful properties?

Possible other useful properties that it could have (optional, should satisfy as many of them as possible):

  1. Min should be associative and commutative.

  2. It should distribute over addition if possible, like in the tropical semiring for real numbers (this is most likely way too strong of a condition but I would be happy to be proven wrong).

  3. Translation invariance: $\min(A + C,B + C) = \min(A,B) + C$ at least when C commutes with A and B (in particular when C is a multiple of the identity matrix). This is somewhat weaker than 4)

  4. $\min(A,B)$ should ideally be continuous in A and B.

  5. the smallest eigenvalue of $\min(A,B)$ should not be "too much smaller" than the smallest eigenvalues of A and B (a strong version of this with equality would follow from 5 and positive definiteness).

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    $\begingroup$ This is an idea for defining $\max$ on positive-definite Hermitian matrices, but if it does work it might be able to be modified into defining $\min$ on positive semi-definite Hermitian matrices. For positive real numbers $x$ and $y$, it's a well-known fact that $\lim_{a \to \infty} (x^a + y^a)^{1/a} = \max\{x,y\}$. These operations all make sense for positive-definite Hermitian matrices, and this should have the right behavior for commuting matrices. $\endgroup$
    – James E Hanson
    Commented Mar 9, 2021 at 10:27
  • $\begingroup$ @JamesHanson The same expression converges to $\min\{x,y\}$ for $a \to -\infty$ (but if the matrices are only semidefinite then the negative exponents are a problem). $\endgroup$
    – Federico Poloni
    Commented Mar 9, 2021 at 11:09
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    $\begingroup$ I was interested in this question a few years ago. Then I found that the set of Hermitian matrices $C$ such that $C\le A$ and $C\le B$ does not admit a largest element ; unless $A$ and $B$ commutte of course. $\endgroup$
    – Denis Serre
    Commented Mar 9, 2021 at 14:20
  • $\begingroup$ Right. The thing with this question is that I'm interested in a tropical semiring-like structure on matrices, so I don't need uniqueness for maximal elements. Right now, a "good enough" choice seems to be $\min(x,y) := \lim_{h \rightarrow 0} \log_h(h^x + h^y)$, due to the ease of proving various properties, particularly translation invariance. $\endgroup$
    – saolof
    Commented Mar 9, 2021 at 14:59
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    $\begingroup$ Consider$$A=\begin{pmatrix}2&1-i\\ 1+i&2\end{pmatrix},\ B=\begin{pmatrix}1&0\\ 0&3\end{pmatrix}$$ For the maximum either $$M=\lim_{n\to\infty}(A^n+B^n)^{1/n}\ \text{or}\ M=\lim_{h\to\infty}\log(e^{hA}+e^{hB})/h$$$$\implies M= \frac12\begin{pmatrix}5+\sqrt{2}&(1-i)+(-1)^{3/4}\\ (1+i)-(-1)^{1/4}& 5+\sqrt{2}\end{pmatrix}$$For the minimum either$$m=\lim_{n\to-\infty}(A^n+B^n)^{1/n}\ \text{or}\ m=\lim_{h\to-\infty}\log(e^{hA}+e^{hB})/h$$ $$\implies m=\frac12\begin{pmatrix}3-\sqrt{2}& (1-i)+(-1)^{3/4} \\ (1+i)-(-1)^{1/4}&3-\sqrt{2}\end{pmatrix}$$Then$$A,B,M,m,A+B-m>0\ \text{but}\ A+B-M\ngeq0$$ $\endgroup$
    – user44143
    Commented Mar 9, 2021 at 17:25

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I don't know if this is the kind of answer that you were looking for, but (hat tip this math.SE answer) the preprint

Nikolas Stott, Maximal lower bounds in the Löwner order [arXiv:1612.05664]

seems to have a detailed discussion of maximal lower bounds of two symmetric matrices with respect to the Löwner order, which is precisely the partial order generated by the cone of positive semi-definite matrices. The conclusion seems to be that maximal lower bounds are generally non-unique (in a partial order a strict maximum may not exist due to incomparable elements), but can be usefully parametrized. Whether there could be a useful way to pick a preferred maximal lower bound by a formula, I can't say.

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  • $\begingroup$ Ah, this definitely adds a lot of information about criterion 2) at least, which is very useful. $\endgroup$
    – saolof
    Commented Mar 9, 2021 at 12:32

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