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\setCJKmainfont[Mapping=tex-text]{PMingLiU}
\setmainfont[Mapping=tex-text]{Times New Roman}
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\usepackage[lite,subscriptcorrection,nofontinfo,zswash]{mtpro2}
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Sunday, February 19, 2012
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 kth 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 kth 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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