User angela - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:17:54Z http://mathoverflow.net/feeds/user/4267 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/57337/when-should-a-supervisor-be-a-co-author/57391#57391 Answer by angela for When should a supervisor be a co-author? angela 2011-03-04T19:32:20Z 2011-03-04T19:32:20Z <p>As a non-mathematician, I am somewhat mystified by the prevailing norms of the mathematics community as I understand them from this thread. Correct me if I'm wrong, but it sounds like: </p> <ul> <li><p>Supervisors make important intellectual contributions to the thesis work of their students. </p></li> <li><p>Typically, the name of the supervisor does not appear on the work. </p></li> </ul> <p>For example, the most upvoted comment at the moment says "as a rule the supervisor should not be a co-author in the main paper taken from a student's thesis, <strong>even if he has contributed substantially to it</strong>." (emphasis is mine) Other comments echo the sentiment. </p> <p>This seems problematic, both morally and practically. In other scientific communities, the author list is supposed to reflect the people who contributed intellectually to the paper. Manipulating it is an ethical offense. For example, <strong>the practices described in this thread appear to violate IEEE policies on authorship</strong> <a href="http://www.ieee.org/documents/opsmanual.pdf" rel="nofollow">which state</a> (Section 8.2.1)</p> <blockquote> <p>Authorship and co-authorship should be based on a substantial intellectual contribution ... the list of authors on an article serves multiple purposes; it indicates who is responsible for the work and to whom questions regarding the work should be addressed.</p> </blockquote> <p>Finally, I would just like to add that as a student, I would feel horrible submitting a paper authored only by me if the paper was based in large part on the insights of someone else. </p> http://mathoverflow.net/questions/40683/which-doubly-stochastic-matrices-can-be-written-as-products-of-pairwise-averaging Which doubly stochastic matrices can be written as products of pairwise averaging matrices? angela 2010-09-30T23:22:23Z 2010-10-01T03:14:39Z <p>A matrix $A$ is called doubly stochastic if its entries are nonnegative, and if all of its rows and columns add up to $1$. A subset of doubly stochastic matrices is the set of pairwise averaging matrices which move two components of a vector closer to their average. More precisely, a pairwise averaging matrix $P_{i,j,\alpha}$ is defined by stipulating that $y=P_{i,j,\alpha}x$ is $$y_i = (1-\alpha) x_i + \alpha x_j$$ $$y_j = \alpha x_i + (1-\alpha) x_j$$ $$y_k = x_k ~~{\rm for~ all }~ k \neq i,j~,$$ where $\alpha \in [0,1]$. </p> <p>My question is: can every doubly stochastic matrix be written as a product of pairwise averaging matrices? </p> <p>If the answer is no, I would like to know if its possible to characterize the doubly stochastic matrices which can be written this way.</p> <p><b>Update:</b> I just realized that the answer is no. Here is a sketch of the proof. Pick any $3 \times 3$ doubly stochastic matrix matrix $A$ with $A_{23}=A_{32}=0$. If $A$ can be written as the product of pairwise averages, the pairwise average matrices $P_{2,3,\alpha}$ never appear in the product, since they result in setting the $(2,3)$ and $(3,2)$ entries to positive numbers, which remain positive after any more applications of pairwise averages. So the product must only use $P_{1,2,\alpha}$ or $P_{1,3,\alpha}$. But one can see that no matter in what order one applies these matrices, at least one of $A_{23}$ or $A_{32}$ will be set to a positive number. For example, if we average 1 and 2 first and then 1 and 3, then $A_{32}$ will be nonzero. </p> <p>My second question is still unanswered: is it possible to characterize the matrices which are products of pairwise averages?</p> http://mathoverflow.net/questions/35618/matrices-self-adjoint-with-respect-to-some-inner-product matrices self-adjoint with respect to some inner product angela 2010-08-15T02:21:42Z 2010-08-15T09:03:51Z <p>Is it possible to give a nice characterization of matrices $A \in R^{n \times n}$ which are self-adjoint with respect to <em>some</em> inner product?</p> <p>These matrices include all symmetric matrices (of course) and some nonsymmetric ones: for example, the transition matrix of any (irreducible) reversible Markov chain will have this property. </p> <p>Naturally, all such matrices must have real eigenvalues, though I do not expect that this is a sufficient condition (is it?).</p> <p>About the only observation I have is that since any inner product can be represented as $\langle x,y \rangle = x^T M y$ for some positive definite matrix $M$, we are looking for matrices $A$ which satisfy $A^T M = M A$ or $M^{-1} A^T M = A$. In other words, we are looking for real matrices similar to their transpose with a positive definite similarity matrix. </p> http://mathoverflow.net/questions/16471/a-geometric-interpretation-of-independence A geometric interpretation of independence? angela 2010-02-26T03:15:47Z 2010-02-26T16:46:50Z <p>Consider the set of random variables with zero mean and finite second moment. This is a vector space, and $\langle X, Y \rangle = E[XY]$ is a valid inner product on it. Uncorrelated random variables correspond to orthogonal vectors in this space. </p> <p>Questions:</p> <p>(i) Does there exist a similar geometric interpretation for independent random variables in terms of this vector space?</p> <p>(ii) A collection of jointly Gaussian random variables are uncorrelated if and only if they are independent. Is it possible to give a geometric interpretation for this?</p> http://mathoverflow.net/questions/57337/when-should-a-supervisor-be-a-co-author/57391#57391 Comment by angela angela 2011-03-04T21:08:41Z 2011-03-04T21:08:41Z @Thierry Zell - Fair enough, but at least in other disciplines the published record is there to set things straight. Based on this discussion, it seems like in mathematics the published record sometimes omits key information. http://mathoverflow.net/questions/57337/when-should-a-supervisor-be-a-co-author/57391#57391 Comment by angela angela 2011-03-04T20:48:14Z 2011-03-04T20:48:14Z @Daniel Litt - I agree. But if standards for authorship in mathematics are so different than in other scientific disciplines, I wish mathematicians had disseminated this fact more widely. To give a concrete example: I have sometimes referred to &quot;the proof of conjecture X by Y,&quot; because there is a paper by mathematician Y which proves conjecture X. I now see that my assumption - that the proof is due solely to the people whose names appear on the paper as authors - might be false. If I want to <i>justly</i> attribute the proof of conjecture X, it appears I need additional information. http://mathoverflow.net/questions/57337/when-should-a-supervisor-be-a-co-author/57391#57391 Comment by angela angela 2011-03-04T20:39:31Z 2011-03-04T20:39:31Z @Andre Henriques - perhaps if mathematicians ordered authors by contribution, supervisors would not feel like they have to remove their names from papers to which they contributed, which frankly sounds to me like it would be an improvement all around.... http://mathoverflow.net/questions/57337/when-should-a-supervisor-be-a-co-author/57391#57391 Comment by angela angela 2011-03-04T20:38:09Z 2011-03-04T20:38:09Z @Neil Strickland - I don't think thats quite correct, authorship information is not presented merely by mentioning the advisor. This is because just mentioning the advisor leaves open the question of whether the advisor made a substantial contribution to the work, enough to be listed as co-author. http://mathoverflow.net/questions/57337/when-should-a-supervisor-be-a-co-author/57391#57391 Comment by angela angela 2011-03-04T20:36:38Z 2011-03-04T20:36:38Z @Mark Meckes - I disagree, I think there is a substantial difference between the two scenarios you compare. In the first scenario, the relative contributions of you and Prof. A are obvious to anyone who reads the literature. In the second scenario, they are hidden. In fact, future histories of the subject are likely emphasize your work and de-emphasize Prof A's, since his/her name is not on any of the published work leading to the proof of the conjecture. http://mathoverflow.net/questions/40683/which-doubly-stochastic-matrices-can-be-written-as-products-of-pairwise-averaging/40702#40702 Comment by angela angela 2010-10-01T02:37:17Z 2010-10-01T02:37:17Z Doesn't Hadamard's inequality ($|det(A)|$ \leq \prod_i ||a_i||) imply that the determinant of every doubly stochastic matrix lies between $−1$ and $1$? http://mathoverflow.net/questions/35468/widely-accepted-mathematical-results-that-were-later-shown-wrong/35654#35654 Comment by angela angela 2010-08-16T03:48:19Z 2010-08-16T03:48:19Z <a href="http://en.wikipedia.org/wiki/Godel" rel="nofollow">en.wikipedia.org/wiki/Godel</a>'s_ontological_proof http://mathoverflow.net/questions/31475/singular-values-of-matrix-sums/31479#31479 Comment by angela angela 2010-07-12T00:24:24Z 2010-07-12T00:24:24Z so if $A=diag(1,0,0)$ and $B=diag(0,1,0)$, then $s_2(A)=s_2(B)=0$, $s_2(A+B)=1$, so the inequality we are discussing gives $1 \leq 0+0$. http://mathoverflow.net/questions/31475/singular-values-of-matrix-sums/31479#31479 Comment by angela angela 2010-07-11T23:46:12Z 2010-07-11T23:46:12Z I don't understand the justification for the inequality $s_k(A+B) \leq s_k(A)+s_k(B)$. The matrices $A=diag(1,0)$, $B=diag(0,1)$ appear to form a counterexample. If you want in addition $k\geq 1$, as implied by your first sentence, then $A=diag(1,0,0)$, $B=diag(0,1,0)$ is a counterexample. http://mathoverflow.net/questions/16471/a-geometric-interpretation-of-independence/16473#16473 Comment by angela angela 2010-02-26T05:02:22Z 2010-02-26T05:02:22Z I am quite happy with answers which add more structure to the space or tinker with the setting in any way. My only motivation is to get some geometric intuition about random variables, so anything in that vein would make me very happy.