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 1d comment Idempotent and nilpotent matrices are defined differently. Why? Idempotent is $A^2=A$ not $A^2=I$. 2d reviewed Edit A Chinese Exam Question which is…quite hard 2d revised A Chinese Exam Question which is…quite hard Corrected grammar and formatting. Apr 29 awarded Nice Question Apr 28 comment Proof of Cayley's Theorem But, I have to say, writing functions on the right is weird in this context when the function is denoting left multiplication... Apr 28 revised Proof of Cayley's Theorem deleted 2 characters in body Apr 28 comment Proof of Cayley's Theorem @rt6 I don't know the OP's reasons, but sometimes I like writing functions on the right because then composition has the nice property that $f \circ g$ means "$f$, then $g$", instead of the reverse. Apr 28 comment What exactly is an “analytic function”? This might be an English sentence parsing issue; in the definition, try replacing 'which is represented' by 'which can be represented'. Apr 21 answered How to succeed in upper-level math Apr 20 comment Quotient by a discrete subgroup of a Lie group If you choose the neighborhoods for the overlaps small enough, then no two points in the neighborhoods will be equivalent under $T$, so you are back to the case of $H$. Apr 19 comment Proof that Gale-Shapley is man optimal The source you found should also define "valid partner" somewhere. A "valid partner" in this context means that two people can be paired up in some stable pairing. It is also sometimes called a "stable partner" or a "possible partner". Maybe the original Gale-Shapley paper will help: cramton.umd.edu/market-design/… Apr 18 comment Scissor equivalence/congruence of two convex hulls How do you get an area of 4 for the second triangle? The triangle is contained in a 2x2 square with area 4 so the triangle has to have smaller area. Apr 18 answered Quotient by a discrete subgroup of a Lie group Apr 13 awarded Nice Answer Apr 13 comment Proof that Gale-Shapley is man optimal How can that be? First, we gave an algorithm (GS) to construct a stable pairing $S^*$. Then, we proved that no man can do better in any other stable pairing than he does in $S^*$. By definition, this means $S^*$ is man-optimal. Apr 13 comment Proof that Gale-Shapley is man optimal Because the argument applies to any pairing $S$ in which some man is better off in $S$ than he is in $S^*$. We prove that any such $S$ is unstable, therefore, $S^*$ is optimal for all men among stable pairings. Apr 12 comment Proof that Gale-Shapley is man optimal Also, you have to have another pairing S involved in the proof because to say that GS is not man optimal means there is another stable pairing in which one of the men has a better partner than in GS. Apr 12 comment Proof that Gale-Shapley is man optimal Your proof does not make sense to me, starting with "Suppose $A$ preferred $Y$ to $Z$". $Y$ is the one getting rejected; the new candidate for $A$ should be someone she prefers over $Y$, not the other way around. But more importantly, in order for there to be unstable pair, both parties should be unhappy with their current partners, so there have to be 4 people involved (the unstable pair + their current partners), which you don't seem to have here. Apr 11 comment Proof that Gale-Shapley is man optimal The proof is showing that any pairing $S$ which purports to be better than $S^*$ (for some man), would have an unstable pair. $S$ has the unstable pair, not $S^*$. $S^*$ is produced by the GS algorithm and is always stable. Apr 10 comment I don't understand what a “free group” is! In simple terms, the group operation is defined by: concatenate the words, then repeatedly cancel consecutive pairs $x_i x_i^{-1}$ or $x_i^{-1} x_i$ until you can't cancel any more. For example $(ab)(b^{-1}c) = ac$. Of course, with this approach, you have to prove that you get the same answer no matter how you cancel.