1. Diagonalization of Linear Operator
Hung-yi Lee

2. Announcement
明天 (5/5, 週四) 助教會來講解作業二
會有 bonus 的作業

3. Independent Eigenvectors
𝑃= 𝑝 1 ⋯ 𝑝 𝑛
𝐷= 𝑑 1 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 𝑑 𝑛
Review
eigenvector
eigenvalue
An n x n matrix A is diagonalizable (𝐴=𝑃𝐷 𝑃 −1 ) =
The eigenvectors of A can form a basis for Rn.
𝑑𝑒𝑡 𝐴−𝑡 𝐼 𝑛
= 𝑡− 𝜆 1 𝑚 1 𝑡− 𝜆 2 𝑚 2 … 𝑡− 𝜆 𝑘 𝑚 𝑘 …… ……
Eigenvalue:
𝜆 1
𝜆 2
𝜆 𝑘 ……
Eigenspace:
Basis for 𝜆 1
Basis for 𝜆 2
Basis for 𝜆 3
Independent Eigenvectors

4. Diagonalization of Linear Operator
𝑇 𝑥 1 𝑥 2 𝑥 3 = 8 𝑥 1 +9 𝑥 2 −6 𝑥 1 −7 𝑥 2 3 𝑥 1 +3 𝑥 2 − 𝑥 3
Example 1:
det -t
𝐴= −6 − −1
The standard matrix is -t -t
 the characteristic polynomial is (t + 1)2(t  2)
 eigenvalues: 1, 2
Eigenvalue -1:
Eigenvalue 2:
B1= − ,
B2= 3 −2 1
 B1  B2 is a basis of R 3
 T is diagonalizable.

5. Diagonalization of Linear Operator
𝑇 𝑥 1 𝑥 2 𝑥 3 = − 𝑥 1 + 𝑥 2 +2 𝑥 3 𝑥 1 − 𝑥 2 0
Example 2:
det -t
𝐴= − −
The standard matrix is -t -t
 the characteristic polynomial is t2(t + 2)
 eigenvalues: 0, 2
 the reduced row echelon form of A  0I3 = A is
1 −
 the eigenspaces corresponding to the eigenvalue 0 has the
dimension 1 < 2 = algebraic multiplicity of the eigenvalue 0
 T is not diagonalizable.

6. This lecture 𝑣 B 𝑇 𝑣 B 𝐵 −1 𝐵 𝑣 𝑇 𝑣 Reference: Chapter 5.4 𝑇 B simple
Properly selected
Properly selected 𝐴 𝑣
𝑇 𝑣

7. Diagonalization of Linear Operator
If a linear operator T is diagonalizable 𝐷
𝑇 B
𝑣 B
𝑇 𝑣 B
simple
Eigenvectors form the good system
𝑃 −1 𝑃
𝐵 −1 𝐵
Properly selected
Properly selected
𝐴=𝑃𝐷 𝑃 −1 𝑣
𝑇 𝑣

8. Diagonalization of Linear Operator
-1: 2:
𝑇 𝑥 1 𝑥 2 𝑥 3 = 8 𝑥 1 +9 𝑥 2 −6 𝑥 1 −7 𝑥 2 3 𝑥 1 +3 𝑥 2 − 𝑥 3
B1= − ,
B2= 3 −2 1
𝑇 B
𝑣 B
𝑇 𝑣 B
− −
− − −1
− −
8 9 0 −6 − −1 𝑣
𝑇 𝑣

9. Another Application of Eigenvector: Page Rank
Reference:
THE $25,000,000,000 EIGENVECTOR: THE LINEAR ALGEBRA BEHIND GOOGLE
/m430/GooglePageRank.pdf
A SURVEY OF EIGENVECTOR METHODS FOR WEB INFORMATION RETRIEVAL