diagonalization - igravigraphics.ics.uci.edu/ics6n/discussion/disc4.pdf · diagonalization theorem...
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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