MIT | 数据分析、信号处理和机器学习中的矩阵方法笔记系列 Lecture 4 Eigenvalues and Eigenvectors

本系列为MIT Gilbert Strang教授的”数据分析、信号处理和机器学习中的矩阵方法”的学习笔记。

Gilbert Strang & Sarah Hansen | Spring 2018
18.065: Matrix Methods in Data Analysis, Signal Processing, and Machine Learning
视频网址: https://ocw.mit.edu/courses/18-065-matrix-methods-in-data-analysis-signal-processing-and-machine-learning-spring-2018/
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矩阵方法

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Lecture 0: Course Introduction

Lecture 1 The Column Space of

A

$A$

Contains All Vectors

Ax

$A x$

Lecture 2 Multiplying and Factoring Matrices

Lecture 3 Orthonormal Columns in

Q

$Q$

Give

′

Q’Q=I

$Q^{'} Q = I$

Lecture 4 Eigenvalues and Eigenvectors

Lecture 5 Positive Definite and Semidefinite Matrices

Lecture 6 Singular Value Decomposition (SVD)

Lecture 7 Eckart-Young: The Closest Rank

k

$k$

Matrix to

A

$A$

Lecture 8 Norms of Vectors and Matrices

Lecture 9 Four Ways to Solve Least Squares Problems

Lecture 10 Survey of Difficulties with

Ax=b

$A x = b$

Lecture 11 Minimizing ||x|| Subject to

Ax=b

$A x = b$

Lecture 12 Computing Eigenvalues and Singular Values

Lecture 13 Randomized Matrix Multiplication

Lecture 14 Low Rank Changes in

A

$A$

and Its Inverse

Lecture 15 Matrices

(

)

A(t)

$A (t)$

Depending on

t

$t$

, Derivative =

dA/dt

$d A / d t$

Lecture 16 Derivatives of Inverse and Singular Values

Lecture 17 Rapidly Decreasing Singular Values

Lecture 18 Counting Parameters in SVD, LU, QR, Saddle Points

Lecture 19 Saddle Points Continued, Maxmin Principle

Lecture 20 Definitions and Inequalities

Lecture 21 Minimizing a Function Step by Step

Lecture 22 Gradient Descent: Downhill to a Minimum

Lecture 23 Accelerating Gradient Descent (Use Momentum)

Lecture 24 Linear Programming and Two-Person Games

Lecture 25 Stochastic Gradient Descent

Lecture 26 Structure of Neural Nets for Deep Learning

Lecture 27 Backpropagation: Find Partial Derivatives

Lecture 28 Computing in Class [No video available]

Lecture 29 Computing in Class (cont.) [No video available]

Lecture 30 Completing a Rank-One Matrix, Circulants!

Lecture 31 Eigenvectors of Circulant Matrices: Fourier Matrix

Lecture 32 ImageNet is a Convolutional Neural Network (CNN), The Convolution Rule

Lecture 33 Neural Nets and the Learning Function

Lecture 34 Distance Matrices, Procrustes Problem

Lecture 35 Finding Clusters in Graphs

Lecture 36 Alan Edelman and Julia Language

Lecture 4: Eigenvalues and Eigenvectors

Last time:

Orthogonal Matrices

This time:

Eigenvectors and Eigenvalues of matrix

Eigenvectors and Eigenvalues of Symmetric Matrices

4.1 Egienvectors and Eigenvalues of matrix A

For
A

x

Ax

$A x$

(

A

$A$

is

×

n

n\times n

$n \times n$

)
- if x is especially chosen well
- Ax就能恰好和x有着
  
  same direction:
- 即**
  
  Ax=\lambda x
  
  $A x = λ x$
  
  **
- Eigenvectors
  $\lambda λ$

Q1:

What is “Eigenvectors” good for?

由定义, 特征向量x很special, 但它useful吗？

special is good, but useful is even better

性质1：关于

k

A^k

$A^{k}$
- Given
  
  x
  
  =
  
  λ
  
  x
  
  Ax = \lambda x
  
  $A x = λ x$
- \Rightarrow
  
  $\Rightarrow$
  
  2
  
  x
  
  A^2 x
  
  $A^{2} x$
  
  =
  
  (
  
  A
  
  x
  
  )
  
  A (Ax)
  
  $A (A x)$
  
  =
  
  (
  
  λ
  
  )
  
  x
  
  A (\lambda)x
  
  $A (λ) x$
  
  =
  
  2
  
  \lambda^2
  
  $λ^{2}$
  
  x
  
  $x$
- x也是
  
  2
  
  A^2
  
  $A^{2}$
  
  的特征向量！
  
  ▪ Eigenvalue:
  
  2
  
  \lambda^2
  
  $λ^{2}$
- 同理，
  
  n
  
  x
  
  =
  
  λ
  
  n
  
  x
  
  A^n x = \lambda^n x
  
  $A^{n} x = λ^{n} x$
  
  ▪
  
  −
  
  1
  
  x
  
  =
  
  1
  
  λ
  
  x
  
  A^{-1} x = \frac{1}{\lambda} x
  
  $A^{- 1} x = \frac{1}{λ} x$
  
  (前提是
  
  ≠
  
  0
  
  \lambda \not ={0}
  
  $λ \neq = 0$
  
  即 A不可逆)
  
  ▪
  
  A
  
  t
  
  x
  
  =
  
  e
  
  λ
  
  t
  
  x
  
  e^{At}x = e^{\lambda t}x
  
  $e^{A t} x = e^{λ t} x$
- Take any vector v
  
  R
  
  n
  
  \in R^n
  
  $\in R^{n}$
  
  ,
  
  ▪ 假设矩阵A有n个不同的特征向量
  
  x
  
  i
  
  =
  
  λ
  
  i
  
  x
  
  i
  
  Ax_i = \lambda_i x_i
  
  $A x_{i} = λ_{i} x_{i}$
  
  (
  
  =
  
  1
  
  ,
  
  2
  
  ,
  
  .
  
  .
  
  .
  
  ,
  
  n
  
  i = 1,2, …, n
  
  $i = 1, 2, . . ., n$
  
  )
  
  ▪ 若v能够被写为这些eigenvectors
  
  i
  
  x_i
  
  $x_{i}$
  
  的combination:
  
  =
  
  c
  
  1
  
  x
  
  1
  
  +
  
  c
  
  2
  
  x
  
  2
  
  +
  
  c
  
  3
  
  x
  
  3
  
  +
  
  .
  
  .
  
  .
  
  +
  
  c
  
  n
  
  x
  
  n
  
  v = c_1 x_1 + c_2 x_2 + c_3 x_3 + … + c_n x_n
  
  $v = c_{1} x_{1} + c_{2} x_{2} + c_{3} x_{3} + . . . + c_{n} x_{n}$
  
  ▪ 则有：
  
  k
  
  v
  
  =
  
  c
  
  1
  
  λ
  
  1
  
  k
  
  x
  
  1
  
  +
  
  c
  
  2
  
  λ
  
  2
  
  k
  
  x
  
  2
  
  +
  
  c
  
  3
  
  λ
  
  3
  
  k
  
  x
  
  3
  
  +
  
  .
  
  .
  
  .
  
  +
  
  c
  
  n
  
  λ
  
  n
  
  k
  
  x
  
  n
  
  A^k v = c_1 \lambda_1^k x_1 + c_2 \lambda_2^k x_2 + c_3 \lambda_3^k x_3 + … + c_n \lambda_n^k x_n
  
  $A^{k} v = c_{1} λ_{1}^{k} x_{1} + c_{2} λ_{2}^{k} x_{2} + c_{3} λ_{3}^{k} x_{3} + . . . + c_{n} λ_{n}^{k} x_{n}$
- 因此，
  
  特征值特征向量的第一个用处：帮助计算
  
  k
  
  x
  
  A^k x
  
  $A^{k} x$
  
  、
  
  A
  
  $A$
  
  的矩阵函数、方程等
  
  ▪ 例如，已知
  
  k
  
  =
  
  A
  
  k
  
  v
  
  V_k = A^k v
  
  $V_{k} = A^{k} v$
  
  , 那么利用Eigenvectors和Eigenvalues, 求
  
  k
  
  +
  
  1
  
  =
  
  A
  
  V
  
  k
  
  V_{k+1} = A V_k
  
  $V_{k + 1} = A V_{k}$
  
  或
  
  V
  
  /
  
  d
  
  t
  
  =
  
  A
  
  v
  
  dV/dt = Av
  
  $d V / d t = A v$
  
  就会很方便
性质2：关于相似矩阵
- Similar matrix: B is similar to A:
  
  ▪
  
  \exist
  
  $\exists$
  
  invertible matrix
  
  :
  
  B
  
  =
  
  M
  
  −
  
  1
  
  A
  
  M
  
  M: B = M^{-1} A M
  
  $M : B = M^{- 1} A M$
- Then, A and B have the same eigenvalues
- i.e.,
  
  Similar matrices
  
  \Rightarrow
  
  $\Rightarrow$
  
  Same Eigenvalues
- MATLAB计算EigenValues的方法：
  
  ▪ 通过多次similar transform: change
  
  A
  
  $A$
  
  \Rightarrow
  
  $\Rightarrow$
  
  1
  
  M_1
  
  $M_{1}$
  
  \Rightarrow
  
  $\Rightarrow$
  
  2
  
  M_2
  
  $M_{2}$
  
  , … ✅
  
  \Rightarrow
  
  $\Rightarrow$
  
  finally to a triangular matrix
  
  ▪ 从而 acquire the eigenvalues
  
  ▪ （If A is a
  
  symmetric matrix
  
  , it will go to a diagonal matrix (不仅仅是triangular matrix)）
- Proof: (Similar matrices
  
  \Rightarrow
  
  $\Rightarrow$
  
  Same Eigenvalues)
  
  ▪ what we know:
  
  =
  
  M
  
  −
  
  1
  
  A
  
  M
  
  B = M^{-1} A M
  
  $B = M^{- 1} A M$
  
  ▪ 若
  
  −
  
  1
  
  A
  
  M
  
  y
  
  =
  
  λ
  
  y
  
  M^{-1} A M y = \lambda y
  
  $M^{- 1} A M y = λ y$
  
  ▪
  
  \Rightarrow
  
  $\Rightarrow$
  
  M
  
  y
  
  =
  
  λ
  
  M
  
  y
  
  AMy = \lambda My
  
  $A M y = λ M y$
  
  ▪
  
  \Rightarrow
  
  $\Rightarrow$
  
  $KaTeX parse error: Undefined control sequence: \lamda at position 10: A (My) = \̲l̲a̲m̲d̲a̲ ̲(My)$
  
  ▪
  
  \Rightarrow
  
  $\Rightarrow$
  
  \lambda
  
  $λ$
  
  也是A的一个 eigenvalue
- 练习1：证明
  
  AB has the same non-zero eigenvalues as BA?
  
  ▪
  
  A
  
  ,
  
  B
  
  ∈
  
  R
  
  n
  
  ×
  
  n
  
  \forall A, B \in R^{n\times n}
  
  $\forall A, B \in R^{n \times n}$
  
  ▪
  
  want
  
  show that AB and BA are similar:
  
  ✅
  
  (
  
  A
  
  B
  
  )
  
  M
  
  −
  
  1
  
  =
  
  B
  
  A
  
  M(AB)M^{-1} = BA
  
  $M (A B) M^{- 1} = B A$
  
  ✅ Just Let
  
  =
  
  B
  
  M = B
  
  $M = B$
  
  #
  
  ✅ 前提是特征值non-zero: 否则cannot determine whether
  
  −
  
  1
  
  M^{-1}
  
  $M^{- 1}$
  
  exists
- 练习2：
  
  矩阵的特征值是否满足可加性、可乘性？
  
  ▪ If I know the eigenvalues
  
  (
  
  A
  
  )
  
  ,
  
  λ
  
  (
  
  B
  
  )
  
  \lambda(A), \lambda(B)
  
  $λ (A), λ (B)$
  
  of
  
  A
  
  $A$
  
  and
  
  B
  
  $B$
  
  , repectively, can I get eigenvalues of AB by
  
  (
  
  A
  
  )
  
  λ
  
  (
  
  b
  
  )
  
  \lambda(A)\lambda(b)
  
  $λ (A) λ (b)$
  
  ?
  
  No!
  
  ▪ also
  
  (
  
  A
  
  )
  
  +
  
  λ
  
  (
  
  B
  
  )
  
  ≠
  
  λ
  
  (
  
  A
  
  +
  
  B
  
  )
  
  \lambda(A) + \lambda(B) \not ={\lambda(A+B)}
  
  $λ (A) + λ (B) \neq = λ (A + B)$

4.2 Egienvectors and Eigenvalues of real Symmetric Matrix S

Are eigenvalues of
S

S

$S$

are real numbers?
- Other real matrices could have imaginary eigenvalues
- Let
  
  S
  
  =
  
  [
  
  0
  
  1
  
  −
  
  1
  
  0
  
  ]
  
  AS = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
  
  $A S = [0 - 1 10]$
  
  ▪
  
  a 90
  
  ^\circ
  
  $^{\circ}$
  
  rotation
  
  ▪ Anit-symmetric matrix
  
  ▪ Are eigenvalues os
  
  S
  
  AS
  
  $A S$
  
  are imaginary numbers
  
  ▪ 物理理解：
  
  90度旋转矩阵AS, 乘以任何一个向量x都不可能不改变该向量的方向
  
  （
  
  x
  
  =
  
  λ
  
  x
  
  Ax = \lambda x
  
  $A x = λ x$
  
  ）
  
  ▪ 所以
  
  特征值
  
  \lambda
  
  $λ$
  
  和特征向量
  
  x
  
  $x$
  
  不是实数而是虚数！
  
  ▪ 数学计算：
  
  ✅
  
  x
  
  =
  
  λ
  
  x
  
  Ax = \lambda x
  
  $A x = λ x$
  
  ✅
  
  \Rightarrow
  
  $\Rightarrow$
  
  存在非0的x，使得
  
  A
  
  −
  
  λ
  
  I
  
  )
  
  x
  
  =
  
  0
  
  (A-\lambda I) x = 0
  
  $(A - λ I) x = 0$
  
  ✅
  
  \Rightarrow
  
  $\Rightarrow$
  
  A
  
  −
  
  λ
  
  I
  
  )
  
  (A-\lambda I)
  
  $(A - λ I)$
  
  不可逆
  
  \Rightarrow
  
  $\Rightarrow$
  
  e
  
  t
  
  (
  
  A
  
  −
  
  λ
  
  I
  
  )
  
  det(A-\lambda I)
  
  $d e t (A - λ I)$
  
  = 0
  
  ✅
  
  e
  
  t
  
  (
  
  A
  
  S
  
  −
  
  λ
  
  I
  
  )
  
  =
  
  [
  
  λ
  
  1
  
  −
  
  1
  
  −
  
  λ
  
  ]
  
  det(AS – \lambda I) = \begin{bmatrix} \lambda & 1 \\ -1 & – \lambda \end{bmatrix}
  
  $d e t (A S - λ I) = [λ - 1 1 - λ]$
  
  =
  
  2
  
  \lambda ^2
  
  $λ^{2}$
  
  + 1 = 0
  
  \Rightarrow
  
  $\Rightarrow$
  
  =
  
  i
  
  ,
  
  −
  
  i
  
  \lambda = i, -i
  
  $λ = i, - i$
  
  \Rightarrow
  
  $\Rightarrow$

To way to check the eigenvalues:

特征值之和 = 对角线之和 (trace)

Add all eigenvalues = Add all diagonal elements

a way to check the calculated eigenvalues

特征值之积 = Determinant (det

A

$A$

)

For Symmetric Matrix:
- eigenvalues are real;
- eigenvectors are orthogonal
  
  ▪ 意味着
  
  full set of eigenvectors
  
  ! (even if some eigenvalues may be repeated)
- check: using
  
  trace
  
  and
  
  det
- Example:
  
  =
  
  [
  
  0
  
  1
  
  1
  
  0
  
  ]
  
  S = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}
  
  $S = [0110]$
  
  ▪ Eigenvalues
  
  =
  
  1
  
  ,
  
  −
  
  1
  
  \lambda = 1, -1
  
  $λ = 1, - 1$
  
  ▪ Eigenvectors
  
  =
  
  [
  
  1
  
  1
  
  ]
  
  ,
  
  [
  
  1
  
  −
  
  1
  
  ]
  
  x = \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix}
  
  $x = [11], [1 - 1]$
  
  ▪
  
  =
  
  [
  
  1
  
  0
  
  0
  
  −
  
  1
  
  ]
  
  \Lambda = \begin{bmatrix} 1 & 0\\ 0 & -1 \end{bmatrix}
  
  $Λ = [10 0 - 1]$
- 特征值相同的矩阵是否相似？
  
  ▪ 上面的
  
  S
  
  $S$
  
  和
  
  \Lambda
  
  $Λ$
  
  have the same eigenvalues
  
  ▪ If they are similar:
  
  \exist
  
  $\exists$
  
  invertible
  
  M
  
  $M$
  
  M
  
  =
  
  M
  
  Λ
  
  SM = M\Lambda
  
  $S M = M Λ$
  
  👉 令
  
  =
  
  [
  
  x
  
  1
  
  ,
  
  x
  
  2
  
  ]
  
  M = [x_1, x_2]
  
  $M = [x_{1}, x_{2}]$
  
  , then
  
  [
  
  x
  
  1
  
  x
  
  2
  
  ]
  
  S \begin{bmatrix} x_1 & x_2\\ \end{bmatrix}
  
  $S [x_{1} x_{2}]$
  
  =
  
  x
  
  1
  
  x
  
  2
  
  ]
  
  [
  
  λ
  
  1
  
  0
  
  0
  
  λ
  
  2
  
  ]
  
  \begin{bmatrix} x_1 & x_2\\ \end{bmatrix} \begin{bmatrix} \lambda_1 & 0\\ 0 & \lambda_2 \end{bmatrix}
  
  $[x_{1} x_{2}] [λ_{1} 0 0 λ_{2}]$
  
  ▪ 看上去是相似的。但是上面的推导有一个前提：
  
  特征值构成的矩阵
  
  M
  
  $M$
  
  必须可逆！
  
  ▪ If Matrix A has eigenvectors
  
  1
  
  ,
  
  x
  
  2
  
  ,
  
  .
  
  .
  
  .
  
  ,
  
  x
  
  n
  
  x_1, x_2, …, x_n
  
  $x_{1}, x_{2}, . . ., x_{n}$
  
  and eigenvalues
  
  1
  
  ,
  
  λ
  
  2
  
  ,
  
  .
  
  .
  
  .
  
  ,
  
  λ
  
  n
  
  \lambda_1, \lambda_2, …, \lambda_n
  
  $λ_{1}, λ_{2}, . . ., λ_{n}$
  
  👉
  
  [
  
  x
  
  1
  
  ,
  
  x
  
  2
  
  ,
  
  .
  
  .
  
  .
  
  ,
  
  x
  
  n
  
  ]
  
  =
  
  [
  
  x
  
  1
  
  ,
  
  x
  
  2
  
  ,
  
  .
  
  .
  
  .
  
  ,
  
  x
  
  n
  
  ]
  
  [
  
  λ
  
  1
  
  .
  
  .
  
  .
  
  0
  
  .
  
  .
  
  .
  
  .
  
  .
  
  .
  
  .
  
  .
  
  .
  
  0
  
  .
  
  .
  
  .
  
  .
  
  .
  
  .
  
  λ
  
  2
  
  ]
  
  A[x_1, x_2, …, x_n] = [x_1, x_2, …, x_n] \begin{bmatrix} \lambda_1 & … & 0\\ … & … & … \\ 0 & … & …\lambda_2 \end{bmatrix}
  
  $A [x_{1}, x_{2}, . . ., x_{n}] = [x_{1}, x_{2}, . . ., x_{n}] ⎣ ⎡ λ_{1} . . . 0 . . . . . . . . . 0 . . . . . . λ_{2} ⎦ ⎤$
  
  👉
  
  \Rightarrow
  
  $\Rightarrow$
  
  X
  
  =
  
  X
  
  Λ
  
  AX = X \Lambda
  
  $A X = X Λ$
  
  🚩
  
  \Rightarrow
  
  $\Rightarrow$
  
  A与
  
  \Lambda
  
  $Λ$
  
  相似：
  
  =
  
  X
  
  Λ
  
  X
  
  −
  
  1
  
  A = X \Lambda X^{-1}
  
  $A = X Λ X^{- 1}$
  
  (
  
  前提是
  
  X
  
  $X$
  
  可逆
  
  )
  
  ✅ 利用上述公式，可直接得到
  
  2
  
  A^2
  
  $A^{2}$
  
  的特征值：
  
  2
  
  =
  
  X
  
  Λ
  
  2
  
  X
  
  −
  
  1
  
  A^2 = X \Lambda^2 X^{-1}
  
  $A^{2} = X Λ^{2} X^{- 1}$
  
  ▪ 对于Symmetric matrix
  
  S
  
  $S$
  
  :
  
  👉 its eigenvectors matrix X is orthogonal: denoted as
  
  Q
  
  $Q$
  
  👉
  
  \Rightarrow
  
  $\Rightarrow$
  
  =
  
  Q
  
  Λ
  
  Q
  
  −
  
  1
  
  =
  
  Q
  
  Λ
  
  Q
  
  T
  
  S = Q \Lambda Q^{-1} = Q \Lambda Q^T
  
  $S = Q Λ Q^{- 1} = Q Λ Q^{T}$
  
  👉
  
  Spectral Theorem
  
  : Every symmetric matrix 都能分解为
  
  Λ
  
  Q
  
  T
  
  Q \Lambda Q^T
  
  $Q Λ Q^{T}$
  
  的形式

原文链接：https://blog.csdn.net/qazwsxrx/article/details/125072212

Lecture 4: Eigenvalues and Eigenvectors

4.1 Egienvectors and Eigenvalues of matrix A

4.2 Egienvectors and Eigenvalues of real Symmetric Matrix S

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