From their website: "Rocky Mountain Journal of Mathematics publishes both research and expository articles in mathematics, and particularly invites well-written survey articles." A search in Math ...

Since free C*-algebras don't exist, we can't give a concrete description of all relations that are allowed. Instead, we need to give axioms that determine what collections of n-tuples $(a_1,\dots,a_n)... View answer 17 votes As jjcale mentions in a comment, the index of a Fredholm operator is very important in physics. One way to define the Chern number of a topological insulator is in terms of the index of a Fredholm ... View answer Accepted answer 15 votes Lin's theorem shows the existence of a localized basis for the low-energy space in models of non-interacting fermions on a finite lattice on a disk. This was observed by Matt Hastings. See "Topology ... View answer 11 votes See Lemma 3.7 in my paper C*-relations, Math. Scand., 107(1), 43--72 (2010). I show that this seminorm is actually a norm. This is for the finite or infinite case. For the case of countably many ... View answer 10 votes Gert Pedersen wrote "In a careless moment a C*-algebraist might be quoted for saying that there is a covariant functor between the categories of commutative C*-algebras with morphisms and the category ... View answer 7 votes You can get a nice estimate using the spectral norm. Theorem 3.2.32 of Bratteli and Robinson vol. 1 tell us an estimate on$f(A) - f(B)$for a variety of functions. Take$f:\mathbb R \rightarrow \...

The Chern numbers, Spin Chern numbers and so forth in condensed matter physics are very important in understanding topological insulators. There are many ways to compute this invariants, and some of ...

For all matrices, even non-invertible matrices, one gets a polar decomposition $X=UP$ with $U$ unitary and $P$ positive semidefinite. Of course $U$ is not unique unless $X$ is invertible. I ...

If you want an explicit projection, you can form a variation of a Rieffel projection as follows. First take any function $f$ from $[-\pi/2,\pi/2]$ to $[0,1]$ that sends $-\pi/2$ to $1$, dips down to $... View answer 5 votes I almost found it: Example 4.4 in "Presentations and Tietze transformations of C*-algebras" by Will Grilliette, New York J. Math. 18 (2012) 121--137. The generator in the concrete algebra is not ... View answer Accepted answer 4 votes For small Hermitian (or real symmetric) matrices, yes, but really this is a hard problem not fully solved. See [1,2] for algorithms. The Cardoso paper  looks at the non-commuting case, but in the ... View answer 3 votes I was looking for this a few years ago, and found it in the appendix of this paper. El-kaïoum, M. Moutuou. "Graded Brauer groups of a groupoid with involution." Journal of Functional Analysis 266.5 (... View answer Accepted answer 3 votes I believe the correct formula is $$(u-1_{A^+}) \otimes p + 1_{(A \otimes B)^+}$$ which makes sense if you work this out in the following special case. Let$A$be$C_0(0,1)$and let$B$be$C(\{0,1\})...

This seems to be an $O(n^3)$ problem, so if your matrix is smaller than 100 by 100 you might as well use an eigensolver. For bigger matrices you can use Newton's method (see Higham's papers or book)...

Determinants are tricky for matrices of quaternions, but they are not as bad when the matrix is hermitian. In that case, one can expand in the usual way along any row and get the same result. See ...

I am rather certain that you cannot modify a standard diagonalization to the structured diagonalization that you want. I have seen such a method used, but it is expected to fail occasionally, ...

Both $\mathrm{id}$ and $\sigma\circ\chi$ map from $A \oplus_C B$ to $A \oplus_C B$ respecting the left and right projection maps. (I leave to the reader the writing out of the diagrams.) Since $A ... View answer 2 votes Have a look at a recent paper discussing how matrix sparseness and locality go together: "Decay Properties of Spectral Projectors with Applications to Electronic Structure" by Benzi et al. in SIAM ... View answer Accepted answer 2 votes Here is a way to answer the second question. I am assuming complex matrices and using '$X^{*}=\overline{X}^{\mathrm{T}}$. The first question (already answered by Suvrit ) I discuss at the end. ... View answer 1 votes I can give you a few papers, and these have references to others. I discuss joint approximate eigenvectors in the context of approximate joint measurement in . You need to know that often it is$...