Diagonalization Hung-yi Lee

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

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

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

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

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

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.

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:

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

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

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!

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 =

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

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, http://mathoverflow.net/questions/51887/non-diagonalizable-doubly-stochastic-matrices在這裡鍵入方程式。 .727 .273 .85 .15 1 0 𝐴=𝑃𝐷 𝑃 −1 𝐴 𝑚 =𝑃 𝐷 𝑚 𝑃 −1

Diagonalizable Diagonalize a given matrix RREF (invertible) RREF

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

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.

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)

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

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

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= 8 9 0 −6 −7 0 3 3 −1 The standard matrix is -t -t  the characteristic polynomial is (t + 1)2(t  2)  eigenvalues: 1, 2 Eigenvalue -1: Eigenvalue 2: B1= −1 1 0 , 0 0 1 B2= 3 −2 1  B1  B2 is a basis of R 3  T is diagonalizable.

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= −1 1 2 1 −1 0 0 0 0 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 −1 0 0 0 1 0 0 0  the eigenspaces corresponding to the eigenvalue 0 has the dimension 1 < 2 = algebraic multiplicity of the eigenvalue 0  T is not diagonalizable.

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

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= −1 1 0 , 0 0 1 B2= 3 −2 1 𝑇 B 𝑣 B 𝑇 𝑣 B −1 0 0 0 −1 0 0 0 2 −1 0 3 1 0 −2 0 1 1 −1 −1 0 3 1 0 −2 0 1 1 8 9 0 −6 −7 0 3 3 −1 𝑣 𝑇 𝑣