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:

• Orthonormal Basis of Hermitian matrices for a Hilbert Space of Operators asks for an orthogonal basis, but elements don't have to be PSD
• Can we construct a basis for Hermitian matrices made of positive semidefinite ones? asks for a basis where elements have to be PSD, but basis doesn't have to be orthogonal (with respect to $trace(A^\dagger B)=0$)
• What is a basis for the space of $n\times n$ Hermitian matrices? asks for any basis of Hermitian matrices whatsoever

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):

• Simultaneous Diagonalizability of Multiple Commuting Matrices
• https://mathoverflow.net/questions/124779/when-are-two-operators-simultaneously-diagonalisable
• Prove that simultaneously diagonalizable matrices commute
• Simultaneously diagonalisable iff A,B commute

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":

• https://en.wikipedia.org/wiki/Biorthogonal_system (maybe this instead is more relevant?)
• section 6.2 of Bridging the gap between general probabilistic theories and the device-independent framework for nonlocality and contextuality, Giulio Chiribella, Xiao Yuan

Cf. also

• Perfectly distinguishable states are mutually orthogonal
• why non orthogonal states are indistinguishable?

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 $N$.