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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


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Q17E

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Found in: Page 840

### Horngren'S Financial And Managerial Accounting

Book edition 6th
Author(s) Tracie L. Miller-Nobles, Brenda L. Mattison
Pages 992 pages
ISBN 9780134486833

# Data for Research Enterprises follows: 201920182017Total current assets$490,000$320,000$230,000Total current liabilities$235,000$160,000$115,000Compute the dollar amount of change and the percentage of change in Research Enterprises’ working capital each year during 2019 and 2018. What do the calculated changes indicate?

The change in working capital in 2019 is $95,000 and in 2018 is$45,000. The percentage change in working capital in 2019 is 59.38% and in 2018 39.13%

See the step by step solution

## Step 4 Indication of the calculations.

The change in working capital is increasing which means the company is expanding their operations so the working capital of the company is increasing.

format: plain


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IFLP   13074
INSTITUTO DE FISICA LA PLATA
Although the magnetism of fine particles has been widely studied during the last years, there are a rich variety of phenomena, which remain to be understood in the nanoscale granular systems. A complete understanding of the magnetic properties of these nanoscopic systems is hindered by their inherent complexity, involving broad size distributions, different structural and/or magnetic phases and interparticle magnetic interactions.  In order to get deeper insight into the nanoestructurated metallic systems, allowed powders of (Fe$_{79}$Mn$_{21}$)$_{1-x}$Cu$_{x}$ (x = 0.1 to 0.35) were prepared by milling the elemental species in a horizontal vibrating device with steel ball and vial, under Ar atmosphere during 15h at 33Hz. The structural analysis has been obtained by X-ray diffraction (XRD) and X-Ray Absorption Fine Structure (EXAFS-XANES) at the Fe K-edge and at the Cu K-edge at RT. Mössbauer spectroscopy (MS) and magnetic measurements (AC-susceptibility, ZFC-FC and M vs H) were performed to magnetic characterized the samples. Preliminary XAFS results show changes in the amplitude of the Fourier transformation. The amplitude increases when decreases the Cu content in the sample. XRD and MS results suggest the presence of two FCC-phases with different ordering temperatures. The AC-susceptibility measurements indicate the existence of a characteristic temperature {\it T$_{c}$} ($\sim$ 150 K) which shifts to lower temperatures with increase the Cu content. These results, associated with magnetization measurements as function of temperature and applied field, provide evidence for a highly disordered magnetic system with spin-glass-like behavior.


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# Math Help - If x goes very large, say infinite, what does this equation will look like

1. ## If x goes very large, say infinite, what does this equation will look like

when x goes very large, say infinite, what does this equation will look like?

H is a constant

1) 1/(1+(x/H)^2)^(1/2))

2) 1/(1+(H/x)^2)^(1/2))

1) 1/x^2

2) 1/x

But I am not too sure how to get there. Thank you

2. Originally Posted by oxxiissiixxo
when x goes very large, say infinite, what does this equation will look like?

H is a constant

1) 1/(1+(x/H)^2)^(1/2))

2) 1/(1+(H/x)^2)^(1/2))

1) 1/x^2

2) 1/x

But I am not too sure how to get there. Thank you

Did you mean?:

$\dfrac{1}{\sqrt{1+(x/H)^2}}$ for 1) ?

3. 1) $\displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(\frac{x}{H}\right)^2}}$

As x grows larger, the denominator of the fraction grows larger. In this regard, the limit is similar to: $\displaystyle \lim_{t\to\infty}\frac{1}{t}$, which is 0.

2) $\displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(\frac{H}{x}\right)^2}}$

As x grows larger, $\frac{H}{x}$ approaches 0. Therefore,

$\displaystyle \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(\frac{H}{x}\right)^2}} = \lim_{x\to\infty} \frac{1}{\sqrt{1+\left(0\right)^2}} = \lim_{x\to\infty} \frac{1}{\sqrt{1+0}} = \lim_{x\to\infty} \frac{1}{\sqrt{1}} = \lim_{x\to\infty} \frac{1}{1} = \lim_{x\to\infty} 1 = 1$


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## Find the Pattern

When attempting to describe a pattern, one approach is to try to find a rule that describes what is happening. If the rule can be applied in every case available to you, it's a valid way to decode the pattern.

What rule describes the pattern above?

There might be many different ways to put what's happening here into words. One way to describe it, though, is this — in the figure above, every $3\times 3$ grid is shaded in the square that corresponds to the position of the $3\times 3$ grid among nine of them.

Two examples — the $3\times 3$ grid in the center has its central square shaded, while the $3\times 3$ grid in the upper left has its small upper-left square shaded.

Given this, we'd expect the bottom-right square to be shaded in the bottom right, and that's the most reasonable way to complete this particular pattern: