So $\det A=\sum_{\pi\in S_n}{\rm sign}(\pi)a_\pi$ where $a_\pi=\prod_i a_{i,\pi(i)}$
If $A$ is skew-symmetric, so $A^T=-A$, then $\det A=\sum_{pi\in ECS_n}{\rm sign}(\pi)a_\pi$ where $ECS_n$ is the set of permutations having only cycles of even length
i was an outlier when i typed my homework as an undergrad in the 1998-2001 time frame. but a number of my fellow students knew latex, they just did not type their homework. in grad school, everyone was all latex, all the time.
i submitted an assignment in law school in latex and got back the remark 'can you do this in word? i need to leave comments on it.' rather than send them a screenshot on how to comment on a PDF, i switched over. i still hate it.
i had to typeset a number of equations in something for work a while ago, and it was a nightmare. although word's equation editor does support a subset of latex.
maybe someone's written a plugin that makes it easier, except anybody with a brain would not be using word, let alone writing plugins to make word less of the hellhole that it is.
my daughter is vegging out in front of the TV for the first time in months. she's watching a documentary about polar bears. she began yelling at the screen when they attacked some seals.
I am not sure if Chineese has it, but German has this thing called Case system which ultimately causes like 4 or 6 words for the same idea of "the". I've heard Russian has much more than that which adds to the complexity
in chinese theres specific words that are paired with specific nouns for counting. so "dog" is "gou狗" and "one" is “yi一" but "one dog" is "yi zhi gou 一只狗”. One book is "yi ben shu 一本书“. Its like the "bowl of" in "one bowl of noodles", but applies to basically all nouns
Every time you swap to rows, the determinant is negated. to the the bottom row to the top takes $n-1$ row swaps. The negative on the $\Delta$ makes that $(-1)^n$. Then, expanding on the top row you get the bottom element of $\Delta$.
@BAYMAX Yes, your $\Delta=\begin{bmatrix}-a\\-b\\-c\end{bmatrix}$ and so my statement is true. $(-1)^3(-c)=c$
Let $S_{\infty}$ be the group of finitely supported permutations of $\Bbb{N}$. I am trying to argue that $S_{\infty}$ is not residually finite. I feel like a stronger claim is true, namely that any homomorphism into a finite group is trivial, but I don't know how to argue this. I could use some hints.
also, I have an inkling suspicion that $\mathrm{sgn}\colon S_{\infty}\rightarrow\mathbb{Z}/2\mathbb{Z}$ might actually be the profinite completion of $S_{\infty}$, but I don't have an argument
If $f$ is a $C^{\infty}$ function and $U$ is a open subset of $R^n$, what does $C^{\infty}(U)$ mean? More generally I've seen $C^{\infty}(C^n)$, $C^{\infty}(R^n)$ etc. Is it the set of all $C^{\infty}$ functions whose domain is $U$?
@Thorgott I think I got it. Because $A_{\infty}$ is an infinite simple group, any finite-index subgroup of $S_{\infty}$ must actually contain $A_{\infty}$. Indeed, if $N$ is a normal finite-index subgroup of $S_{\infty}$, then $|A_{\infty} : A_{\infty} \cap N| \le |S_{\infty} : N| < N$. Infinite simple groups do not have proper finite-index subgroups, so $A_{\infty} = A_{\infty} \cap N \le N$.
Hence, the intersection of all finite-index normal subgroups of $S_{\infty}$ will contain $A_{\infty}$ which is certainly not trivial.
So, $S_{\infty}$ is not residually finite. This means there exists an element $g \in S_{\infty}$ which is killed by every group homomorphism from $S_{\infty}$ to a finite group...interesting...I wonder what that element looks like.
i can define a gradient of a functional $f$ on a Hilbert space $H$ via reisz representation theorem like this right : given $f:H\to\mathbb{R}$, the Frechet derivative of $f$ is an operator $Df:H\to \mathbb{R}$, i.e $Df\in H'$, so by reisz representation we can associate to $Df$ a unique $v\in H$ such that $Df(u)=\langle u, v \rangle $. We call $v$ the gradient of $f$.
