幾天前科大 reject 了我 mphil 的申請，而我亦沒有報讀其他的院校，半年前在想 334 應該會需要更多的人手負責教學，應該沒問題。但我期後聽某些教授說因為 "productivity" 之類的資源 ``計算" 方法，PGS (Postgraduate Studentship) 被削減了，收生也不比上年多。
完了...，沒有 offer，沒有申請任何工作，也沒有考任何語文試，半年後的我會是怎樣呀......。難得辛辛苦苦二年學了不少知識 (第一年比較渾渾噩噩)，好像都白讀了。可能半年後的我不會再接觸數學........。
Sunday, March 25, 2012
Monday, March 12, 2012
哈哈...
科 number theory 走向變成 algebra course 的路線。上一堂講到對於 finite field $ F$，若 $ f\in F[X]$ 係 irreducible，$ F[X]/(f)$ 同樣係 finite field。因為以前冇乜點諗過呢類 example，開頭諗都係諗返有 division algorithm 云云，同覺得幾神奇，咁就證到 $ (f)$ 係 maximal。但其實我己經學過喺 PID 裏面所有 prime ideal 都係 maximal ...，而 $ (f)$ prime $ \iff$ $ f$ prime (i.e., irreducible in UFD) ...。學完又係唔記得！
Friday, March 2, 2012
Vocabulary
Cavalier (adj) 不經思考 (e.g. we shouldn't be completely cavalier in doing something)
In the sequel = In the following
haphazard (adj) 無計劃地 (it is not easy to haphazardly guess a functor that ...)
trump (verb) 勝過 e.g.:
You should always follow the style guide given to you by a particular publisher or company. That trumps everything else.
ironclad (adj) 不可打破、不可變動
I know there’re no ironclad rules regarding this subject.
dread to do something 不敢去想做某事 (因太可怕)
have yet to do something 仍未做某事
stubborn (adj.) 頑固
compromise (n.) 妥協方案
intricate (adj.) 難以理解
miraculously (adj.) 神奇地
gratuitous (adj.) 無必要的
e.g. The proof is gratuitously complicated.
honestly (adv.) 老實講
In the sequel = In the following
haphazard (adj) 無計劃地 (it is not easy to haphazardly guess a functor that ...)
trump (verb) 勝過 e.g.:
You should always follow the style guide given to you by a particular publisher or company. That trumps everything else.
ironclad (adj) 不可打破、不可變動
I know there’re no ironclad rules regarding this subject.
dread to do something 不敢去想做某事 (因太可怕)
have yet to do something 仍未做某事
stubborn (adj.) 頑固
compromise (n.) 妥協方案
intricate (adj.) 難以理解
miraculously (adj.) 神奇地
gratuitous (adj.) 無必要的
e.g. The proof is gratuitously complicated.
honestly (adv.) 老實講
Sunday, February 19, 2012
Record some code
Use xeCJK and mtpro2 at the same time:
\documentclass[cm-default,no-math]{article}
\usepackage[BoldFont,SlantFont]{xeCJK}
\setCJKmainfont[Mapping=tex-text]{PMingLiU}
\setmainfont[Mapping=tex-text]{Times New Roman}
\usepackage[scaled=0.92]{helvet}
\usepackage[lite,subscriptcorrection,nofontinfo,zswash]{mtpro2}
\straightbraces
\documentclass[cm-default,no-math]{article}
\usepackage[BoldFont,SlantFont]{xeCJK}
\setCJKmainfont[Mapping=tex-text]{PMingLiU}
\setmainfont[Mapping=tex-text]{Times New Roman}
\usepackage[scaled=0.92]{helvet}
\usepackage[lite,subscriptcorrection,nofontinfo,zswash]{mtpro2}
\straightbraces
Tuesday, February 14, 2012
Material of presentation for SVD
Prepared a draft for my presentation:
http://ihome.ust.hk/~cclee/document/PRESENTSVD.pdf
The proof in the original text is too short and seemingly incomplete to me, so I modified the proof and filled in many details. Let me first describe terminology used in this notes, and let me for the moment describe everything in $ \mathbb R$ instead of $ \mathbb C$. Let $ \mathbb{R}^{m\times n}$ denote the collection of $ m\times n$ dimensional real matrices. For $ x\in \mathbb R^k$, write $ x = (x^1,x^2,\dots,x^k)$ (i.e., $ x^k$ is the $ k$th coordinate of $ x$), define $ \|x\|_2=\sqrt{\sum_{i=1}^k |x^i|^2}$.
We also define the $ (n-1)$-sphere to be $ S^{n-1}:=\{x\in \mathbb R^n:(x^1)^2+\cdots+(x^n)^2=1\}$. For $ A\in \mathbb R^{m\times n}$ we define the operator norm
\[ \|A\|_2= \sup_{x\in \mathbb R^n, \|x\|_2=1}\|Ax\|_2.\]
We assume, and what we are trying to prove in my draft, that $ AS^{n-1}=\{Ax:x\in S^{n-1}\}$ is a ``hyperellipse" (or just an ellipse when $ n \leq 3$). Suppose also that $ m\ge n$ and $ A$ is of full rank, then it is legitimate to assume there are unit vectors $ u_1,\dots,u_n$ which point in the direction of semi-axises of the hyperellipse.
We can describe a general construction for $ u_i$'s: first find a vector in $ AS^{n-1}$ that has largest length, call it $ a\in \mathbb R^{m}$, then find another vector $ b$ in $ AS^{n-1} \cap (\mathbb R a)^\perp$ that has largest length, then find another such vector in $ AS^{n-1} \cap (\mathbb R a)^\perp\cap (\mathbb R b)^\perp$ and continue the process until we find $ n$ such vectors. This is indeed how we find semi-axises of an ellipse in $ \mathbb R^3$ and $ \mathbb R^2$.
Now in the way above, the length of semi-axis in the direction of $ u_i$, $ \sigma_i$, is in nonincreasing order, i.e., $ \sigma_1\ge \sigma_2\ge\cdots\ge \sigma_n>0$. Take the unit vector $ v_i \in A^{-1}(\sigma_i u_i)$, one gets \[A\underbrace{\left[\begin{array}{c|c|c}v_1&\cdots&v_n\end{array}\right]}_{:=V}=\left[\begin{array}{c|c|c}u_1&\cdots&u_n\end{array}\right]\begin{bmatrix}\sigma_1&& \\ &\ddots & \\ & & \sigma_n\end{bmatrix}.\]
Let $ u_{n+1},\dots, u_m$ be orthonormal basis in $ (\mathrm{span}_{\mathbb R}\{u_1,\dots,u_n\})^\perp$, then RHS of the above equation becomes
\[\underbrace{\left[\begin{array}{c|c|c}u_1&\cdots&u_m\end{array}\right]}_{:= U}\underbrace{\left[\begin{array}{ccc}\sigma_1&& \\ &\ddots & \\ & & \sigma_n \\ \hline &\mathcal O& \end{array}\right]}_{:=\Sigma},\]
where $ \mathcal O$ denotes a matrix with only 0 entries. One can rewrite the above as: $ AV = U\Sigma$. It will be proved in my draft that indeed $ \{v_i\}$ is orthonormal, hence we can conclude both $ V,U$ are unitary, and we arrive to the expression $ A=U\Sigma V^*$. Owing to this decomposition those $ u_i$'s are called left-singular vector, $ v_i$'s are called right-singular vector and ``diagonal" elements (i.e., if $ \Sigma=(d_{ij})$, diagonal elements are $ d_{ii}$'s) are called singular values. These basically are all the motivation of the general result in my draft.
http://ihome.ust.hk/~cclee/document/PRESENTSVD.pdf
The proof in the original text is too short and seemingly incomplete to me, so I modified the proof and filled in many details. Let me first describe terminology used in this notes, and let me for the moment describe everything in $ \mathbb R$ instead of $ \mathbb C$. Let $ \mathbb{R}^{m\times n}$ denote the collection of $ m\times n$ dimensional real matrices. For $ x\in \mathbb R^k$, write $ x = (x^1,x^2,\dots,x^k)$ (i.e., $ x^k$ is the $ k$th coordinate of $ x$), define $ \|x\|_2=\sqrt{\sum_{i=1}^k |x^i|^2}$.
We also define the $ (n-1)$-sphere to be $ S^{n-1}:=\{x\in \mathbb R^n:(x^1)^2+\cdots+(x^n)^2=1\}$. For $ A\in \mathbb R^{m\times n}$ we define the operator norm
\[ \|A\|_2= \sup_{x\in \mathbb R^n, \|x\|_2=1}\|Ax\|_2.\]
We assume, and what we are trying to prove in my draft, that $ AS^{n-1}=\{Ax:x\in S^{n-1}\}$ is a ``hyperellipse" (or just an ellipse when $ n \leq 3$). Suppose also that $ m\ge n$ and $ A$ is of full rank, then it is legitimate to assume there are unit vectors $ u_1,\dots,u_n$ which point in the direction of semi-axises of the hyperellipse.
We can describe a general construction for $ u_i$'s: first find a vector in $ AS^{n-1}$ that has largest length, call it $ a\in \mathbb R^{m}$, then find another vector $ b$ in $ AS^{n-1} \cap (\mathbb R a)^\perp$ that has largest length, then find another such vector in $ AS^{n-1} \cap (\mathbb R a)^\perp\cap (\mathbb R b)^\perp$ and continue the process until we find $ n$ such vectors. This is indeed how we find semi-axises of an ellipse in $ \mathbb R^3$ and $ \mathbb R^2$.
Now in the way above, the length of semi-axis in the direction of $ u_i$, $ \sigma_i$, is in nonincreasing order, i.e., $ \sigma_1\ge \sigma_2\ge\cdots\ge \sigma_n>0$. Take the unit vector $ v_i \in A^{-1}(\sigma_i u_i)$, one gets \[A\underbrace{\left[\begin{array}{c|c|c}v_1&\cdots&v_n\end{array}\right]}_{:=V}=\left[\begin{array}{c|c|c}u_1&\cdots&u_n\end{array}\right]\begin{bmatrix}\sigma_1&& \\ &\ddots & \\ & & \sigma_n\end{bmatrix}.\]
Let $ u_{n+1},\dots, u_m$ be orthonormal basis in $ (\mathrm{span}_{\mathbb R}\{u_1,\dots,u_n\})^\perp$, then RHS of the above equation becomes
\[\underbrace{\left[\begin{array}{c|c|c}u_1&\cdots&u_m\end{array}\right]}_{:= U}\underbrace{\left[\begin{array}{ccc}\sigma_1&& \\ &\ddots & \\ & & \sigma_n \\ \hline &\mathcal O& \end{array}\right]}_{:=\Sigma},\]
where $ \mathcal O$ denotes a matrix with only 0 entries. One can rewrite the above as: $ AV = U\Sigma$. It will be proved in my draft that indeed $ \{v_i\}$ is orthonormal, hence we can conclude both $ V,U$ are unitary, and we arrive to the expression $ A=U\Sigma V^*$. Owing to this decomposition those $ u_i$'s are called left-singular vector, $ v_i$'s are called right-singular vector and ``diagonal" elements (i.e., if $ \Sigma=(d_{ij})$, diagonal elements are $ d_{ii}$'s) are called singular values. These basically are all the motivation of the general result in my draft.
Saturday, February 11, 2012
嗯...
最近發覺和不會主動接觸任何 pure math 的 (UG) applied math 學生相處是一件非常困難的事。我認為十分 concrete 和自然的 example 他們也會感動反感，abstract 一點 (實際上也不抽像...) 他們會感到不實際，沒有意義。難道 applied math 只須學習 compute，compute 和 compute 嗎？費解。
我做 presentation，多說一點，多舉一些例子，只務求聽眾能學得深入。抽像的東西不一定是難，我反而覺得 concrete 的東西很多時更難。學會抽像的部分能加深對 concrete case 的了解。讀 algebra 時就是很多 concrete case 更需要思考呀...。
我做 presentation，多說一點，多舉一些例子，只務求聽眾能學得深入。抽像的東西不一定是難，我反而覺得 concrete 的東西很多時更難。學會抽像的部分能加深對 concrete case 的了解。讀 algebra 時就是很多 concrete case 更需要思考呀...。
Saturday, February 4, 2012
Confirmed Enrollment
應該唔會有大變動：
CIVL 1160 - Civil Engg & Society
HUMA 1710 - Art of Thinking in HK Context
MATH 3426 - Sampling
MATH 4221 - Euclid & Non-Euclid Geom
MATH 4822A - Intro to Modern Num Theory
MATH 4981D - COMP Linear Algebra
MATH 5112 - Advanced Algebra II
除某二個 4 credit math course 外其他都係 3 credit。因為今個 sem 有幾科 math course 都係冇 tutorial，所以讀 7 科但時間表都唔係太密。
HUMA 1710 - Art of Thinking in HK Context
MATH 3426 - Sampling
MATH 4221 - Euclid & Non-Euclid Geom
MATH 4822A - Intro to Modern Num Theory
MATH 4981D - COMP Linear Algebra
MATH 5112 - Advanced Algebra II
除某二個 4 credit math course 外其他都係 3 credit。因為今個 sem 有幾科 math course 都係冇 tutorial，所以讀 7 科但時間表都唔係太密。
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