Can we construct an ORTHOGONAL ($trace(A^\dagger B) = 0$) basis for Hermitian matrices made of PSD (positive semi-definite) Hermitian matrices?

The space of $$N \times N$$ Hermitian matrices can be construed as a real vector space with dimension $$N^2$$. One can define an inner product on this real vector space by $$\langle A, B \rangle := trace(A^\dagger B)$$. Thus Hermitian matrices $$A$$ and $$B$$ are orthogonal if and only if $$trace (A^\dagger B) = 0$$. The Hermitian matrix $$A$$ is positive semidefinite (PSD) if and only if its eigenvalues are all non-negative.

Now I know it is possible to choose $$N^2$$ linearly independent PSD Hermitian matrices, which thus span the real vector space. Via Gram-Schmidt we can turn this into $$N^2$$ orthogonal Hermitian matrices. But Gram-Schmidt requires taking differences, and in general the difference of two PSD Hermitian matrices need not again be PSD.

So a priori there does not seem to be any guarantee that all of the resulting $$N^2$$ orthogonal Hermitian matrices will still be PSD. Maybe it's always the case that at most $$M < N^2$$ remain PSD, so an orthogonal and PSD basis is impossible?

Question: What is the maximal number of orthogonal ($$trace(A^\dagger B) = 0$$) PSD Hermitian matrices?

Is it $$N^2$$? Or is it e.g. $$N$$?

(We can construct an orthogonal basis of PSD Hermitian matrices if and only if this number is $$N^2$$.)

Motivation: I am not sure, but if the maximal number is less than $$N^2$$ I think it might be related to the claims made e.g. here and here that the "information dimension" of a quantum system is $$N$$ while the vector space dimension of the state space is $$N^2$$. In particular, the definition of "perfectly distinguishable states" seems to require (at least in the quantum case) matrices which are both PSD and orthogonal to each other w.r.t. some real-bilinear form. Because it should be possible to go from any bilinear form to the "standard" one via an invertible transformation, it would seem the maximal number of perfectly distinguishable states should equal the maximal number of orthogonal PSD matrices. Maybe.

Related questions:

EDIT 2: This other question on Math.SE actually might be equivalent? Show that if $\sum_{j=1}^m E_j=I$ and $E_j \geq 0$ then the dual basis cannot be positive semi-definite. Anyway the answers to both are concordant

Update after Answer - Related questions (simultaneous diagonalization):

This Arxiv preprint says (cf. part (ii) of Lemma 2) that the book

R. A. Horn and C. R. Johnson. Matrix analysis. Cambridge University Press, Cambridge, 1985

gives a proof of the equivalence of pairwise commuting and simultaneous diagonizability. (EDIT: it is Theorem 1.3.21 in the 2nd edition, 2013.)

Apparently the argument about being perfectly distinguishable being equivalent to orthogonality with respect to some bilinear form is actually correct, the key term apparently being "biorthogonal":

Cf. also

This preprint (by same authors of one of the preprints above, also published in Physical Review A behind paywall) says:

" In quantum theory a set of distinguishable states {ρi}ni=1 is a set of density [PSD] matrices with orthogonal support. An example of discriminating test for this set is the collection of orthogonal projectors {Pi }ni=1 , where Pi is the projector on the support of ρi for all i<n, while Pn =I− n−1Pi. Clearly, i=1 the maximum number of distinguishable states available for a certain system is the dimension d of the corresponding Hilbert space. In this case, the distinguishable states are rank-one projectors on an orthonormal basis, and the corresponding discriminating test is the projective measurement on the same basis."

The largest number of elements that a collection of orthogonal positive semidefinite nonzero $$N\times N$$ matrices can have is $$N$$.
First note the following facts. For two positive semidefinite $$N\times N$$ matrices $$A$$ and $$B$$, one has the following equivalence: $$AB=0 \quad\iff\quad \langle A,B\rangle=0.$$ Moreover, for positive semidefinite matrices $$A$$ and $$B$$, it holds that $$AB=0$$ if and only if $$BA=0$$. Thus, $$\langle A,B\rangle =0$$ implies that $$AB-BA=0$$, hence the matrices $$A$$ and $$B$$ commute.
Now, if $$\{A_1,A_2,\dots,A_m\}$$ is any collection of pairwise orthogonal, positive semidefinite matrices (i.e., $$\langle A_i,A_j\rangle =0$$ for all $$i,j\in\{1,2,\dots,m\}$$ such that $$i\neq j$$), it must be the case that they are all simultaneously commuting. Because they are pairwise commuting, this collection is simultaneously diagonalizable. Thus, we may suppose without loss of generality that $$A_1,A_2,\dots,A_m$$ are all diagonal matrices. The dimension of the span of all diagonal matrices is $$N$$, so the span of $$\{A_1,\dots,A_m\}$$ is at most $$N$$.
• To confirm, $AB=0$ if and only if $BA=0$ because $AB = 0$ $\implies$ $trace(AB)=0$ $\implies$ $trace(BA)=0$ $\implies$ $BA=0$? And then the key fact (I think) is that "a pairwise commuting collection of (Hermitian?) matrices is simultaneously diagonalizable"? (The simultaneously being important because obviously they are all "individually" diagonalizable due to the spectral theorem.) That is a new fact for me, so I wanted to make sure I understood correctly. Nov 29, 2021 at 23:40
• Anyway I am really glad that the answer to this question was a lot easier than I had thought -- I was afraid I would never find out. The reasoning for this answer sounds correct because I have read in several places that "pure" (rank-1/projection) states are perfectly distinguishable if and only if (considered as linear operators) they have orthogonal supports -- the argument about simultaneous diagonizability explains how our "restriction gets moved from an $N^2$-D space to an $N$-D space" (i.e. the domain of the operator/the space of the vector $v$ so that the operator equals $vv^\dagger$). Nov 30, 2021 at 0:01
• You're right about the "Hermitian" part being important. The generalized version of the simultaneously-commuting fact is the following: Let $\mathcal{A}$ be a family of diagonalizable $n\times n$ matrices. Then $\mathcal{A}$ is a commuting family (i.e., all elements of $\mathcal{A}$ are pairwise commuting) if and only if all elements of $\mathcal{A}$ are simultaneously diagonalizable. There are many references for this fact, but when I googled it this is the first thing that came up (if you want a proof). math.uh.edu/~bgb/Courses/Math6304/MatrixTheory-20120911.pdf Dec 1, 2021 at 2:28
• And yes, your reasoning for $AB=0$ $\iff$ $BA=0$ is correct. Dec 1, 2021 at 2:44
• Finally (to touch on your discussion of distinguishability), a family of density matrices $\{\rho_1,\dots,\rho_m\}$ is perfectly distinguishable (i.e., if you are given one of these state there is a measurement you can perform to perfectly distinguish this sate) if and only if the states are pairwise orthogonal. Thus, the largest number of perfectly distinguishable states is $N$. Dec 1, 2021 at 2:48