因為感覺到自己的 algebra 底十分十分弱，所以這個 winter break 的時間都用來磨練一下自己的 algebra。遇到過的 proof 也盡可能自己做一次。有一些重點想記下 (也為了方便回顧)，所以就以 bullet point 形式打了份 notes。
http://ihome.ust.hk/~cclee/document/notes_on_algebra.pdf (notes 不斷更新，鏈結地址不變)
不是所有 proof 都會記下。太長 (e.g. 所有 field 都可以嵌入一個 algebraically closed field) 的不會記下。若證明只需數行又或是雖然非常長但證明技巧有機會再用到的都會記下 (有時用來引證自己記下的重點)。
Monday, January 16, 2012
Tuesday, December 20, 2011
Independent study/Sping course list
presentation 的 "講義" (?)
http://ihome.ust.hk/~cclee/document/Inde_Presentation.pdf
蠻費精神的。第二個 section 有點無謂，kin li 早在 MATH370 把所有 Hilbert space 的 dual 都找出來了...。我想做 L^1 的 dual 比較有意義？
**********
spring course list 已出，已決定 MATH5112, 4321, 4822A, 4221, HUMA 1710, CENG 1700.
http://ihome.ust.hk/~cclee/document/Inde_Presentation.pdf
蠻費精神的。第二個 section 有點無謂，kin li 早在 MATH370 把所有 Hilbert space 的 dual 都找出來了...。我想做 L^1 的 dual 比較有意義？
**********
spring course list 已出，已決定 MATH5112, 4321, 4822A, 4221, HUMA 1710, CENG 1700.
Friday, December 9, 2011
Saturday, December 3, 2011
無聊地想 linear algebra & independent study 的 presentation
大概在上一個星期的 numerical analysis (MATH3312/231) 課 professor 毛毛 提到了一個 fact：
\sigma(A):=\{\lambda\in\mathbb{C}:A-\lambda I\text{ is not inveritble}\}
\] 為 $ A$ 的 spectrum (也就是說 $ \sigma(A)$ 是 $ A$ 所有的 eigenvalue)。且定義 spectral radius 為 $ \rho(A)=\max\{|\lambda|:\lambda \in \sigma(A)\}$。
上面 $ \lim_{n\to\infty}A^n=0$ 可理解為 $ \lim_{n\to\infty}A^n \mathbf x = 0,\forall \mathbf x\in \mathbb{R}^n$。Google 一下 spectral radius 這個字就有以上那個結果的證明，且用到了 Jordan decomposition，當然我有一個更簡單的解 (TA 也確定是正確的證明)，不須用到這工具。但看來就是沒有人有興趣~，還有人說這是 trivial 的 (那是一位 applied math 學生)，我無言了。
我們現在證明上述定理。可知 $(\Rightarrow)$ 是顯然的，現證 $(\Leftarrow)$：
Proof. It is a well-known result in linear algebra that every matrix can be made into an upper triangular matrix under a change of basis. That is, by reviewing $A$ as a linear operator from $\C^n$ to $\C^n$ rather than a real matrix, there must be a basis in $\C^n$ so that the matrix of $A$ w.r.t. this basis becomes upper triangular. A simple proof can be found in ALGEBRA written by Michael Artin.
We upper triangularize $A$ by some $P\in GL_n(\C)$, i.e.,
\[U:=PAP^{-1}=
\begin{bmatrix}
\lambda_1& b_{12} &\cdots & b_{1n} \\
0 &\lambda_2& \cdots & b_{2n} \\
0 & 0 & \ddots & \vdots \\
0 &0 & \cdots &\lambda_n
\end{bmatrix},
\] then $U\mathbf e_1=\lambda_1 \mathbf e_1$, and $U^j \mathbf e_1=\lambda_1^j\mathbf e_1$ and hence $|\lambda_1|<1\implies U^k \mathbf e_1\to 0$.
We complete the proof by induction, assume there is $k\in \N$ so that
\[\lim_{j\to\infty} U^j(\mathbf e_1),\dots,\lim_{j\to\infty} U^j(\mathbf e_{k-1})=0.\] Then since $U(\mathbf e_{k})=\sum_{i=1}^{k-1} b_{ik}\mathbf e_i + \lambda_k \mathbf e_k$, we have
\[U^{j+1}(\mathbf e_{k})=\sum_{i=1}^{k-1} b_{ik}U^{j}(\mathbf e_i) + \lambda_k U^j(\mathbf e_k).\] For a vector $v\in \C^n$ let's denote $v^\ell$ to be its $\ell^\text{th}$ coordinate, then one has
\[ (U^{j+1}(\mathbf e_{k}))^\ell=\sum_{i=1}^{k-1} b_{ik}(U^{j}(\mathbf e_i))^\ell + \lambda_k (U^j(\mathbf e_k))^\ell,\] this implies
\[ \big|(U^{j+1}(\mathbf e_{k}))^\ell\big|\leq\sum_{i=1}^{k-1} |b_{ik}|\big|(U^{j}(\mathbf e_i))^\ell\big| + |\lambda_k| \big|(U^j(\mathbf e_k))^\ell\big|. \] By induction hypothesis $\lim_{j\to\infty}U^j (\mathbf e_i)=0$ for $i=1,2,\dots,k-1$, one has for any $\epsilon>0$, there is an $N$ such that \[
j>N\implies \big|(U^{j+1}(\mathbf e_{k}))^\ell\big|<\epsilon+ |\lambda_k| \big|(U^j(\mathbf e_k))^\ell\big|.
\] This actually implies $\lim_{k\to\infty} \big|(U^{j+1}(\mathbf e_{k}))^\ell \big|=0$ since $|\lambda_k|<1$. As this is true for $\ell=1,2,\dots,n$, so $\lim_{j\to\infty} U^j(\mathbf e_k)=0$.
We conclude by induction that $\lim_{j\to\infty} U^j(\mathbf e_k)=0$ for $k=1,2,\dots, n$. Since each vector in $\C^n$ is spanned by $\{\mathbf e_i\}_{i=1}^n$, we conclude $U^j\to 0$. Since $A^j=P^{-1}U^j P$, we conclude $A^j\to 0$ on $\C^n$, and of course, on $\R^n\subseteq \C^n$.$\qed$
Remark. For the ($\Leftarrow$) we may also use a famous result in Banach algebra. It is known that the formula of spectral radius of the matrix $A$ is $\limn \|A^n\|^{1/n}$. So $\limn \|A^n\|^{1/n} < 1$ implies we can choose $r<1$ such that there is $N\in\N$, \[
n>N\implies \|A^n\| < r^n \implies \limn \|A^n\|=0
.\] Finally for each $x\in \R^n$, $\|A^n x\|\leq \|A^n\| \|x\|$, so $A^n \to 0$ pointwise on $\R^n$.
話說我在這個 sem 跟 kin li 學 Fourier analysis。初頭也是正正常常地學 fourier series on $ \mathbb{R}/\mathbb{Z}$，其後因為我在修讀的 MATH5111/511 教 finite group representation 的關係，我便開始讀 noncommutative 的 harmonic analysis 了...。最後我的 presentation project 更差點變成 existence of Haar measure on locally compact metrizable group。我其後在想，這和 fourier analysis 關係不太大吧...，便要求只證 existence of Haar measure on compact group，取而代之以 fourier series 的方法 determine $ L^2(Q,m)$ 的 dual，其中 $ Q\subseteq\mathbb{R}$ 是 measurable 及 $ m$ 是 Lebesgue measure。
Theorem. Let $ A$ be real $ n\times n$ matrix, then $\lim_{n\to\infty}A^n=0\iff\rho(A)<1$.其中當 $A$ 為 $n\times n$ real matrix，我們定義 \[
\sigma(A):=\{\lambda\in\mathbb{C}:A-\lambda I\text{ is not inveritble}\}
\] 為 $ A$ 的 spectrum (也就是說 $ \sigma(A)$ 是 $ A$ 所有的 eigenvalue)。且定義 spectral radius 為 $ \rho(A)=\max\{|\lambda|:\lambda \in \sigma(A)\}$。
上面 $ \lim_{n\to\infty}A^n=0$ 可理解為 $ \lim_{n\to\infty}A^n \mathbf x = 0,\forall \mathbf x\in \mathbb{R}^n$。Google 一下 spectral radius 這個字就有以上那個結果的證明，且用到了 Jordan decomposition，當然我有一個更簡單的解 (TA 也確定是正確的證明)，不須用到這工具。但看來就是沒有人有興趣~，還有人說這是 trivial 的 (那是一位 applied math 學生)，我無言了。
我們現在證明上述定理。可知 $(\Rightarrow)$ 是顯然的，現證 $(\Leftarrow)$：
Proof. It is a well-known result in linear algebra that every matrix can be made into an upper triangular matrix under a change of basis. That is, by reviewing $A$ as a linear operator from $\C^n$ to $\C^n$ rather than a real matrix, there must be a basis in $\C^n$ so that the matrix of $A$ w.r.t. this basis becomes upper triangular. A simple proof can be found in ALGEBRA written by Michael Artin.
We upper triangularize $A$ by some $P\in GL_n(\C)$, i.e.,
\[U:=PAP^{-1}=
\begin{bmatrix}
\lambda_1& b_{12} &\cdots & b_{1n} \\
0 &\lambda_2& \cdots & b_{2n} \\
0 & 0 & \ddots & \vdots \\
0 &0 & \cdots &\lambda_n
\end{bmatrix},
\] then $U\mathbf e_1=\lambda_1 \mathbf e_1$, and $U^j \mathbf e_1=\lambda_1^j\mathbf e_1$ and hence $|\lambda_1|<1\implies U^k \mathbf e_1\to 0$.
We complete the proof by induction, assume there is $k\in \N$ so that
\[\lim_{j\to\infty} U^j(\mathbf e_1),\dots,\lim_{j\to\infty} U^j(\mathbf e_{k-1})=0.\] Then since $U(\mathbf e_{k})=\sum_{i=1}^{k-1} b_{ik}\mathbf e_i + \lambda_k \mathbf e_k$, we have
\[U^{j+1}(\mathbf e_{k})=\sum_{i=1}^{k-1} b_{ik}U^{j}(\mathbf e_i) + \lambda_k U^j(\mathbf e_k).\] For a vector $v\in \C^n$ let's denote $v^\ell$ to be its $\ell^\text{th}$ coordinate, then one has
\[ (U^{j+1}(\mathbf e_{k}))^\ell=\sum_{i=1}^{k-1} b_{ik}(U^{j}(\mathbf e_i))^\ell + \lambda_k (U^j(\mathbf e_k))^\ell,\] this implies
\[ \big|(U^{j+1}(\mathbf e_{k}))^\ell\big|\leq\sum_{i=1}^{k-1} |b_{ik}|\big|(U^{j}(\mathbf e_i))^\ell\big| + |\lambda_k| \big|(U^j(\mathbf e_k))^\ell\big|. \] By induction hypothesis $\lim_{j\to\infty}U^j (\mathbf e_i)=0$ for $i=1,2,\dots,k-1$, one has for any $\epsilon>0$, there is an $N$ such that \[
j>N\implies \big|(U^{j+1}(\mathbf e_{k}))^\ell\big|<\epsilon+ |\lambda_k| \big|(U^j(\mathbf e_k))^\ell\big|.
\] This actually implies $\lim_{k\to\infty} \big|(U^{j+1}(\mathbf e_{k}))^\ell \big|=0$ since $|\lambda_k|<1$. As this is true for $\ell=1,2,\dots,n$, so $\lim_{j\to\infty} U^j(\mathbf e_k)=0$.
We conclude by induction that $\lim_{j\to\infty} U^j(\mathbf e_k)=0$ for $k=1,2,\dots, n$. Since each vector in $\C^n$ is spanned by $\{\mathbf e_i\}_{i=1}^n$, we conclude $U^j\to 0$. Since $A^j=P^{-1}U^j P$, we conclude $A^j\to 0$ on $\C^n$, and of course, on $\R^n\subseteq \C^n$.$\qed$
Remark. For the ($\Leftarrow$) we may also use a famous result in Banach algebra. It is known that the formula of spectral radius of the matrix $A$ is $\limn \|A^n\|^{1/n}$. So $\limn \|A^n\|^{1/n} < 1$ implies we can choose $r<1$ such that there is $N\in\N$, \[
n>N\implies \|A^n\| < r^n \implies \limn \|A^n\|=0
.\] Finally for each $x\in \R^n$, $\|A^n x\|\leq \|A^n\| \|x\|$, so $A^n \to 0$ pointwise on $\R^n$.
話說我在這個 sem 跟 kin li 學 Fourier analysis。初頭也是正正常常地學 fourier series on $ \mathbb{R}/\mathbb{Z}$，其後因為我在修讀的 MATH5111/511 教 finite group representation 的關係，我便開始讀 noncommutative 的 harmonic analysis 了...。最後我的 presentation project 更差點變成 existence of Haar measure on locally compact metrizable group。我其後在想，這和 fourier analysis 關係不太大吧...，便要求只證 existence of Haar measure on compact group，取而代之以 fourier series 的方法 determine $ L^2(Q,m)$ 的 dual，其中 $ Q\subseteq\mathbb{R}$ 是 measurable 及 $ m$ 是 Lebesgue measure。
Wednesday, October 12, 2011
Classification
最近在 (midterm 前一日) 溫習 midterm 時不忘抽時間完成 MATH511 的 take-home midterm，最近令我痛苦萬分的莫過於 classify 一個給定 order 的 group。我們的 midterm 問題都是從 Artin (2nd edition) 裏抽出來，第 6, 7 課合共 10 題，其中一題就是 classify group of order 33, 18, 20, 30。33 和 30 還好，但對初初起手去做 classification 的我來說， 18 及 20 是一埸悲劇。你要做的，和 classify order 12 (書的其中一個 subsection) 沒兩樣...。
Friday, October 7, 2011
Link:
Chapter 5 不錯看！
http://www.math.ucdavis.edu/~hunter/book/pdfbook.html
作者也有一些 measure theory 的 notes，嗯........，有空才看。
http://www.math.ucdavis.edu/~hunter/book/pdfbook.html
作者也有一些 measure theory 的 notes，嗯........，有空才看。
Sunday, October 2, 2011
note to myself
- For $ 2L$ periodic functions, $ \displaystyle a_n = \frac{1}{L}\int_{-L}^Lf(x)\cos nx\,dx$ for $ \displaystyle n\ge 1$ and $ \displaystyle b_n=\frac{1}{L}\int_{-L}^L f(x)\sin nx\,dx$ for $ \displaystyle n\ge 1$.
- $ \displaystyle\hat f(0)=\frac{a_0}{2}$ and for $ \displaystyle n\ge 1$, $ \displaystyle \hat f(n) = \frac{a_n- i b_n}{2}$, $ \displaystyle \hat f(-n) = \frac{a_n+i b_n}{2}$, which is easily shown by expanding $ \displaystyle \frac{a_0}{2} + \sum_{k=1}^n (a_k\cos kx + b_k\sin kx)$.
where $ \displaystyle \hat f := (\hat f(0),\hat f(1),\hat f(-1),\hat f(2),\hat f(-2),\dots)\in \ell^2$ has the same norm as that of $ \displaystyle f$. Thus we say that the linear map $ \displaystyle f\mapsto \hat f:L^2(\mathbb T)\to \ell^2$ is isometric. Ok, let's prove that this is equivalent to (for real $ f\in L^2[-L,L]$) \[ \displaystyle\frac{1}{2L} \int_{-L}^{L} |f|^2 = \frac{(a_0)^2}{4}+\sum_{n=1}^\infty \frac{a_n^2+b_n^2}{2}.\]
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