I take issue with the article on Linear Logic in the nLab. It says:

(In this way, linear logic has a perfect de Morgan duality.) The logical rules for negation can then be proved.

While this is true for *Classical* Linear Logic, for Intuitionistic Linear Logic things are very different and I feel that this whole 'dark side of the moon' is being thrown away for no good reason. There is no reason to decide that Classical Logic is better than Intuitionistic Logic and there is not reason to decide that Classical Linear Logic is better than Intuitionistic Linear Logic. the article should have a discussion of both realms. How do I go about adding this discussion, please?

thanks

Valeria ]]>

Cycling is not a sort of breakdown that the Simplex algorithm occasionally does, but the sign of the existence of extraordinary solutions that assign positive values to more than $m$ decision-variables, where $m$ is the number of our structural constraints. Simplex, by its very nature, cannot find such solutions but is able to emit a signal of their existence by cycling.

Consider the classical example given by E. M. L. Beale\footnote{The coefficient of $x_{5}$ on the objective function is altered slightly for reasons that will soon be understood.} where $n=7$ and $m=3$ [see Bazaraa et al., p.176]:

$\begin{array}{lr} max.& & & &+\frac{3}{4}x_{4} & -15x_{5} & +\frac{1}{2} x_{6} & - 6x_{7} & RHS \\ ST: &+x_{1} & & &+\frac{1}{4} x_{4} & -8x_{5} & -x_{6} & +9x_{7} & = 0 \\ & &+x_{2} & & +\frac{1}{2} x_{4} & -12x_{5} & -\frac{1}{2} x_{6} & +3x_{7} & = 0 \\ & & &+x_{3} & & & +x_{6} & & = 1 \end{array}$$\qquad \qquad$ and nonnegativity constraints.

As we all know, Simplex algorithm can assign positive values to at most three (basic) decision variables out of seven of this problem. Now let $\eta$ be a positive number. Then $x_{4}=\frac{2}{5}+\frac{112}{5}\eta$, $x_{5}=0+\eta$, $x_{6}=1$ and $x_{7}= \frac{1}{10}+\frac{4}{15} \eta$ is such an extraordinary solution that assigns positive values to four variables! The objective function accordingly assumes the unbounded value $(1+ \eta)/5$. Even if the third scarce resource is not tapped (i.e. $x_{3}=1$), it assumes the value $(0+\eta)/5$ which is again as big as we wish. Hadn’t we modified the objective function coefficient of $x_{5}$, it would come out as $-\infty$ in which case cycling must really be avoided.

This outcome is categorically different from the “unbounded solutions” we are familiar with. Let’s define a “sub-basis” as the set of current basis’ column vectors but the “departing” one. We already know something about such bases. For instance each row of the current basis’ multiplicative inverse (discernible as a specific part of a specific row of the current Tableau) is the outward-normal (gradient) of the subspace spanned by such a sub-basis. In every cycling example given by Gass et al., and notably in the non-trivial one due to Nering and Tucker (p.310), there is a subspace spanned by such a sub-basis and perpendicular to the RHS vector! Moreover, if such a sub-basis can be extended to a basis that spans the associated subspace positively, then, in the case of a Product Mix Problem for instance, the manufacturer has attained some self-sufficiency to produce at least one product whose production is restrained by intangible constraints only. In the Portfolio Management Problem, the investor’s “credibility”.

\textbf{References}

\renewcommand*\labelenumi{[\theenumi]}

\begin{enumerate} \item Bazaraa, M. S., Jarvis, J. J. \& Sherali, H. D., Linear Programming and Network Flows, Copyright \copyright 2010 by John Wiley \& Sons Inc. \item Beale, E. M. L., Cycling in the Dual Simplex Algorithm, Naval Research Logistics Quarterly 2(4), pp.269-276 December 1955 \item Gass, S. I. \& Vinjamuri, S., Cycling in linear programming problems, Computers \& Operations Research 31 (2004) 303-311 \end{enumerate}

]]>http://iml.univ-mrs.fr/~girard/TS2.pdf

also check out 'Transcendental Syntax and GoI: interplays between logic and philosophy' - V.M. Abrusci, P. Pistone (the link to the paper is broken but the HTML version works on google scholar) http://scholar.google.co.uk/scholar?cluster=2835623818545577955&hl=en&as_sdt=0,5 ]]>

I hope this is an appropriate use of the forum. I’d like to find out what proportion of mathematicians are aware of a certain fact, and I’m not feeling in the mood for asking at the Café, so I thought I’d try here.

Let $V$ and $W$ be finite-dimensional vector spaces over an algebraically closed field. Let $f: V \to W$ and $g: W \to V$ be linear maps. Then $g f$ and $f g$ have the same eigenvalues, with the same algebraic multiplicities, with the possible exception of 0.

It’s not terribly hard to prove this fact. But the question is: did you already know this, off the top of your head?

(I can say something about why I’m asking, but I’d rather wait until I’ve got a few answers.)

Thanks!

]]>I’m struggling to further develop the page on Schur functors, which Todd and I were building. But so far I’ve only done a tiny bit of polishing. I deleted the discussion Todd and I were having near the top of the page, replacing it by a short warning that the definition of Schur functors given here needs to be checked to see if it matches the standard one. I created a page on linear functor and a page on tensor power, so people could learn what those are. And, I wound up spending a lot of time polishing the page on exterior algebra. I would like to do the same thing for tensor algebra and symmetric algebra, but I got worn out.

In that page, I switched Alt to $\Lambda$ as the default notation for exterior algebra. I hope that’s okay. I think it would be nice to be consistent, and I think $\Lambda$ is most widely used. Some people prefer $\bigwedge$.

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