<![CDATA[Category: Metase | Blog 1]]> 2017-01-28T23:37:00+08:00 https://vincenttam.github.io/ Octopress <![CDATA[A Comment for New Users on Math.SE]]> 2017-01-10T19:34:53+08:00 https://vincenttam.github.io/blog/2017/01/10/a-comment-for-new-users-on-math-dot-se Background

Mathematics learners ask questions on Mathematics Stack Exchange to get an answer from others without geographical or time restrictions, unlike teachers in schools. Users answer questions to gain virtual points called reputation.

## Problem

Among them, many are new users who don’t use MathJax while typing their questions. As a result, the output is difficult to read. This discourages users from answering those questions, so we would have less answers to read. Since we can sometimes benefit from alternative solutions, it’s better to post a question that is clear enough to attract others to offer an answer. We can downvote those poorly rendered questions, but I think it’s a bit cruel to do so on the very first post from new users. Therefore, I choose to leave a comment which suggests them to use $\rm \LaTeX$. They often say they don’t know how to use it. To avoid responses like that, I include a link to the MathJax guide on Meta Math.SE.

Please use $\rm \LaTeX$.

However, it’s tedious to type the markdown source code every time I want to leave this comment.

## Solution

I’ll include the code below, so that it can be simply copied and pasted next time.

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<![CDATA[Avoided a Duplicate Question]]> 2017-01-07T02:05:10+08:00 https://vincenttam.github.io/blog/2017/01/07/avoided-a-duplicate-question Problem

I encountered the following math problem, so I typed in on Mathematics Stack Exchange. Then, a list of similar posts appeared.

Let $E = {1,\dots,p}$, where $p$ is a prime number. $G$ is a transitive subgroup of $S_p$, and $H$ is a nontrivial normal subgroup of $G$. Show that $H$ acts transitively on $E$.

I notice that $\lvert G \cdot x \rvert = \lvert E \rvert = p$ because $G$ acts transitively on $E$. Now, I try to show the same for $H$, but I am stuck at $\lvert H \cdot x \rvert$. May I say that for all $g \in G$, $\lvert gH \cdot x \rvert = \lvert H \cdot x \rvert$? Why can’t $H$ fix $x$?

Is it a possible duplicate of another question?

If the answer is yes, then my post is going to be quickly flagged as duplicate by experienced users on this site, especially those who are strong at abstract algebra, and I’ll lose reputation for that.

I’ve really found the same problem on Mathematics Stack Exchange by luck.

## Lessons learnt

I viewed this question four days ago. When I tried to access it again tonight from my web browsing history, I found this a bit hard. Therefore, as a post on Mathematics Meta stack Exchange suggests, Vote early, vote often.

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<![CDATA[A Failed Review ★]]> 2016-12-31T23:03:38+08:00 https://vincenttam.github.io/blog/2016/12/31/a-failed-review The problem is of the same type of the one described in Try to comment: Fail review audit on Meta Mathematics Stack Exchange. As a result, I’m not going to post my problem again so as not to create another duplicate question.

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<![CDATA[Simplify a Determinant Identity]]> 2016-12-29T03:40:58+08:00 https://vincenttam.github.io/blog/2016/12/29/simplify-a-determinant-identity Background

Suppose that $A$ and $B$ are singular and nonsingular matrices respectively. Simplify $\det((A+B)^2−(A−B)^2)$.

## Problem

A wrong solution with a vote of -2 is chosen by Daniel. Why can this happen?

## Possible explanation

That’s because he’s correctly done the expansion until $\det(2AB + 2BA)$.

## Raison d’être of this post

Having spent time on typing a comment, I worry that it will automatically disappear in sooner or later if the accepted answer is deleted. Therefore, I back it up here.

Consider $% <![CDATA[ A = \begin{bmatrix} 1&2&3\\ 0&0&0\\ 5&7&9 \end{bmatrix} , B = \begin{bmatrix} 3&2&3\\ 2&2&1\\ 3&1&3 \end{bmatrix}. %]]>$ % <![CDATA[ \begin{align*} \det(AB+BA) &= \det\left( \begin{bmatrix} 16&9&14 \\ 0&0&0 \\ 56&33&49 \end{bmatrix} +\begin{bmatrix} 18&27&36 \\ 7&11&15 \\ 18&27&36 \end{bmatrix} \right) \\ &=\det\left( \begin{bmatrix} 34&36&50\\ 7&11&15\\ 74&60&85 \end{bmatrix} \right)= 30 \ne 0. \end{align*} %]]> However, if $A = 0$ and $B = I_3$, then the answer is clearly zero. As a result, we can’t decude further from $\det(2(AB + BA))$.

## Lessons learnt

The generation of a random matrix/array of integers using randi([imin, imax], m, n). For more details, you may read GNU Octave’s manual.

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