1. Diagonalization
Hung-yi Lee

2. Review If 𝐴𝑣=𝜆𝑣 (𝑣 is a vector, 𝜆 is a scalar)
𝑣 is an eigenvector of A
𝜆 is an eigenvalue of A that corresponds to 𝑣
Eigenvectors corresponding to 𝜆 are nonzero solution of (A  In)v = 0
A scalar 𝑡 is an eigenvalue of A
excluding zero vector
Eigenvectors corresponding to 𝜆
Eigenspace of 𝜆:
Eigenvectors corresponding to 𝜆 + 𝟎
=𝑁𝑢𝑙𝑙(A  In)  𝟎
eigenspace
𝑑𝑒𝑡 𝐴−𝑡 𝐼 𝑛 =0

3. Review Characteristic polynomial of A is 𝑑𝑒𝑡 𝐴−𝑡 𝐼 𝑛
Factorization
multiplicity
= 𝑡− 𝜆 1 𝑚 1 𝑡− 𝜆 2 𝑚 2 … 𝑡− 𝜆 𝑘 𝑚 𝑘 ……
Eigenvalue:
𝜆 1
𝜆 2
𝜆 𝑘
Eigenspace:
𝑑 1
𝑑 2
𝑑 𝑘
(dimension)
≤ 𝑚 1
≤ 𝑚 2
≤ 𝑚 𝑘

4. Outline An nxn matrix A is called diagonalizable if 𝐴= 𝑃𝐷 𝑃 −1
D: nxn diagonal matrix
P: nxn invertible matrix
Is a matrix A diagonalizable?
If yes, find D and P
Reference: Textbook 5.3
Application 1: growth
Application 2: linear operator

5. Diagonalizable An nxn matrix A is called diagonalizable if 𝐴= 𝑃𝐷 𝑃 −1
D: nxn diagonal matrix
P: nxn invertible matrix
Not all matrices are diagonalizable
(?)
A2 = 0
If A = PDP1 for some invertible P and diagonal D
A2 = PD2P1
= 0
D2 = 0
D = 0
D is diagonal
A= 0

6. Diagonalizable 𝑃= 𝑝 1 ⋯ 𝑝 𝑛 𝐷= 𝑑 1 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 𝑑 𝑛
𝑃= 𝑝 1 ⋯ 𝑝 𝑛
Diagonalizable
𝐷= 𝑑 1 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 𝑑 𝑛
If A is diagonalizable
𝐴=𝑃𝐷 𝑃 −1
𝐴𝑃=𝑃𝐷
𝐴𝑃= 𝐴 𝑝 1 ⋯ 𝐴 𝑝 𝑛
𝑃𝐷=𝑃 𝑑 1 𝑒 1 ⋯ 𝑑 𝑛 𝑒 𝑛
= 𝑃 𝑑 1 𝑒 1 ⋯ 𝑃 𝑑 𝑛 𝑒 𝑛
= 𝑑 1 𝑃 𝑒 1 ⋯ 𝑑 𝑛 𝑃 𝑒 𝑛
= 𝑑 1 𝑝 1 ⋯ 𝑑 𝑛 𝑝 𝑛
𝐴 𝑝 𝑖 = 𝑑 𝑖 𝑝 𝑖
𝑝 𝑖 is an eigenvector of A corresponding to eigenvalue 𝑑 𝑖

7. Diagonalizable = = = 𝑃= 𝑝 1 ⋯ 𝑝 𝑛 𝐷= 𝑑 1 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 𝑑 𝑛
𝑃= 𝑝 1 ⋯ 𝑝 𝑛
Diagonalizable
𝐷= 𝑑 1 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 𝑑 𝑛
If A is diagonalizable
𝑝 𝑖 is an eigenvector of A corresponding to eigenvalue 𝑑 𝑖
𝐴=𝑃𝐷 𝑃 −1 =
There are n eigenvectors that form an invertible matrix =
There are n independent eigenvectors =
The eigenvectors of A can form a basis for Rn.

8. Diagonalizable 𝑃= 𝑝 1 ⋯ 𝑝 𝑛 𝐷= 𝑑 1 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 𝑑 𝑛
𝑃= 𝑝 1 ⋯ 𝑝 𝑛
Diagonalizable
𝐷= 𝑑 1 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 𝑑 𝑛
If A is diagonalizable
𝑝 𝑖 is an eigenvector of A corresponding to eigenvalue 𝑑 𝑖
𝐴=𝑃𝐷 𝑃 −1
How to diagonalize a matrix A?
Is diagonalizable unique? No.
The eigen values can repeat in step 2.
Find n L.I. eigenvectors corresponding if possible, and form an invertible P
Step 1:
The eigenvalues corresponding to the eigenvectors in P form the diagonal matrix D.
Step 2:

9. Diagonalizable
A set of eigenvectors that correspond to distinct eigenvalues is linear independent.
𝑑𝑒𝑡 𝐴−𝑡 𝐼 𝑛
Factorization
= 𝑡− 𝜆 1 𝑚 1 𝑡− 𝜆 2 𝑚 2 … 𝑡− 𝜆 𝑘 𝑚 𝑘 …… ……
Eigenvalue:
𝜆 1
𝜆 2
𝜆 𝑘 ……
Eigenspace:
𝑑 1
≤ 𝑚 1
𝑑 2
≤ 𝑚 2
𝑑 𝑘
≤ 𝑚 𝑘
(dimension)
Independent

10. Diagonalizable
A set of eigenvectors that correspond to distinct eigenvalues is linear independent.
𝜆 1
𝜆 2
Eigenvalue:
𝜆 𝑚 ……
Assume dependent
𝑣 𝑚
Eigenvector:
𝑣 1
𝑣 2 ……
a contradiction
vk = c1v1+c2v2+  +ck1vk1
Avk = c1Av1+c2Av2+  +ck1Avk1
(k)
kvk = c11v1+c22v2+  +ck1k1vk1 -
kvk = c1kv1+c2kv2+  +ck1kvk1
0 = c1(1k) v1+c2(2k)v2+  +ck1(k1k)vk1
Not c1 = c2 =  = ck1 = 0
Same eigenvalue
a contradiction

11. Independent Eigenvectors
𝑃= 𝑝 1 ⋯ 𝑝 𝑛
Diagonalizable
𝐷= 𝑑 1 ⋯ 0 ⋮ ⋱ ⋮ 0 ⋯ 𝑑 𝑛
If A is diagonalizable
𝑝 𝑖 is an eigenvector of A corresponding to eigenvalue 𝑑 𝑖
𝐴=𝑃𝐷 𝑃 −1
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
Is diagonalizable unique? No.
The eigen values can repeat in step 2. ……
Eigenspace:
Basis for 𝜆 1
Basis for 𝜆 2
Basis for 𝜆 3
Independent Eigenvectors
You can’t find more!

12. Diagonalizable - Example
Diagonalize a given matrix
characteristic polynomial is (t + 1)2(t  3)
eigenvalues: 3, 1
eigenvalue 3
A = PDP1, where
B1 =
eigenvalue 1
B2 =

13. Application of Diagonalization
If A is diagonalizable,
Example:
𝐴=𝑃𝐷 𝑃 −1
𝐴 𝑚 =𝑃 𝐷 𝑚 𝑃 −1
.15
Study FB
.85
.97
.03
.85 ……
.85
Study
Study
Study
.15
.727
.85
.15
.03 …… FB FB
.97
.273
.15

14. From Study FB To Study FB =𝐴 …… …… .727 .273 .85 .15 1 0 𝐴=𝑃𝐷 𝑃 −1
.03
.85
.97 =𝐴
.85 ……
.85
Study
Study
Study
.15
.727
.85
.15
.03 …… FB FB
.97
.273
.15
The following example shows that stochastic matrices do not need to be diagonalizable,
1 0
𝐴=𝑃𝐷 𝑃 −1
𝐴 𝑚 =𝑃 𝐷 𝑚 𝑃 −1

15. Diagonalizable
Diagonalize a given matrix
RREF
(invertible)
RREF

16. Application of Diagonalization
When 𝑚→∞,
The beginning condition does not influence.
𝐴 𝑚 = 1/6 1/6 5/6 5/6

17. Test for a Diagonalizable Matrix
An n x n matrix A is diagonalizable if and only if both the following conditions are met.
The characteristic polynomial of A factors into a product of linear factors.
𝑑𝑒𝑡 𝐴−𝑡 𝐼 𝑛
Factorization
= 𝑡− 𝜆 1 𝑚 1 𝑡− 𝜆 2 𝑚 2 … 𝑡− 𝜆 𝑘 𝑚 𝑘 ……
For each eigenvalue l of A, the multiplicity of l equals the dimension of the corresponding eigenspace.

18. Independent Eigenvectors
An n x n matrix A is diagonalizable =
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:
𝑑 1
= 𝑚 1
𝑑 2
= 𝑚 2
𝑑 𝑘
= 𝑚 𝑘
(dimension)

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

20. 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

21. 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.

22. 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.

23. 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

24. 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