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？
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spring course list 已出，已決定 MATH5112, 4321, 4822A, 4221, HUMA 1710, CENG 1700.
Tuesday, December 20, 2011
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。
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