Diagonalization

• Let 𝐴 = 𝑃𝐷𝑃−1 and compute 𝐴4

Diagonalization

• Let 𝐴 = 𝑃𝐷𝑃−1 and compute 𝐴4

• 𝐴 = 𝑃𝐷𝑃−1 ⇒ 𝐴4 = 𝑃𝐷𝑃−1 𝑃𝐷𝑃−1 𝑃𝐷𝑃−1 𝑃𝐷𝑃−1

• 𝐴4 = 𝑃𝐷(𝑃−1𝑃)𝐷(𝑃−1𝑃)𝐷(𝑃−1𝑃)𝐷𝑃−1

• = 𝑃𝐷4𝑃−1

• 𝑃𝐷4𝑃−1 =5 72 3

16 00 1

3 −7−2 5

Diagonalization

• Example: The matrix A is factored in the form 𝑃𝐷𝑃−1. Use the Diagonalization Theorem to find the eigenvalues of A and a basis for eigenspace.

• From diagonalization theorem the eigenvalues are the diagonal elements of D.

• While eigenvectors are columns of P.

Diagonalization

• Diagonalize the following matrices. The real eigenvalues are given.

• First find eigenvectors

• 𝐴 − 𝜆𝐼 𝑥 = 0

• 𝜆 = 1

• 𝐴 − 1𝐼 𝑥 =1 2 −11 2 −1−1 −2 1

•1 21 2−1 −2

−1 0−1 01 0

⇒1 20 00 0

−1 00 00 0

• X2 and x3 -> free variables −2𝑠 + 𝑡

𝑠𝑡

→−210

•−2𝑠 + 𝑡

𝑠𝑡

• basis for first eigenspace−210

,101

• Same we calculate second eigenvector

• Second eigenvector will be −1−11

• 𝐷 =1 0 00 1 00 0 5

𝑃 =−2 1 −11 0 −10 1 1

Change of Variable

• Find the change of variable 𝑥 = 𝑃𝑦 that transforms the quadratic form 𝑥𝑇𝐴𝑥 into 𝑦𝑇𝐷𝑦.

Type equation here.

• The matrix of the quadratic form on the left is

• Eigenvalues of A are 9, 6 and 3.

• Then by we can calculate eigenvectors.

• 𝐴 − 𝜆𝐼 𝑥 = 0

• After normalizing eigenvectors we have:

• The desired change of variable is 𝑥 = 𝑃𝑦• Each column of P is one of the eigenvectors.

• 𝑥1 =1

3𝑦1 +

2

3𝑦2 +

−2

3𝑦3

• 𝑥2 =2

3𝑦1 +

1

3𝑦2 +

2

3𝑦3

• 𝑥3 =−2

3𝑦1 +

2

3𝑦2 +

1

3𝑦3