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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.