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Lect26

# Lect26 - Chapter 6 Eigenvalues Eigenvectors Diagonalization...

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Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Eigenvalues & Eigenvectors Method Method Eigenvalues : Solve det ( A - λ I ) = 0 for λ Eigenvector : Find nonzero vectors in the nullspace N ( A - λ I ) Note that det ( A - λ I ) is a polynomial in λ and we call p ( λ ) = det ( A - λ I ) the characteristic polynomial of A . det ( A - λ I ) = 0 is called a characteristic equation .

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Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Eigenvalues & Eigenvectors Examples Example . Find the eigenvalues & the corresponding eigenvectors of A = 3 2 3 - 2 .
Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Eigenvalues & Eigenvectors Examples Example . Find the eigenvalues & the corresponding eigenvectors of A = 3 2 3 - 2 . Ans . The characteristic equation is 3 - λ 2 3 - 2 - λ = 0 . i.e. λ 2 - λ - 12 = 0 . So λ = 4 or - 3 .

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Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Eigenvalues & Eigenvectors Examples Example . Find the eigenvalues & the corresponding eigenvectors of A = 3 2 3 - 2 . Ans . The characteristic equation is 3 - λ 2 3 - 2 - λ = 0 . i.e. λ 2 - λ - 12 = 0 . So λ = 4 or - 3 . Solving ( A - 4 I ) x = 0 , we get x = α ( 2 1 ) T . Any nonzero multiple of 2 1 is an eigenvector belonging to the eigenvalue 4. Eigenvector belonging to the eigenvalue - 3 : β - 1 3 , β = 0
Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Eigenvalues & Eigenvectors (Cont’d) Examples Example . Find the eigenvalues & the corresponding eigenvectors of B = 2 - 3 1 1 - 2 1 1 - 3 2 .

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Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Eigenvalues & Eigenvectors (Cont’d) Examples Example . Find the eigenvalues & the corresponding eigenvectors of B = 2 - 3 1 1 - 2 1 1 - 3 2 . Ans . The characteristic equation is 2 - λ - 3 1 1 - 2 - λ 1 1 - 3 2 - λ = 0 . i.e. - λ ( λ - 1 ) 2 = 0 . So λ = 0 or 1 .
Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Eigenvalues & Eigenvectors (Cont’d) Examples Example . Find the eigenvalues & the corresponding eigenvectors of B = 2 - 3 1 1 - 2 1 1 - 3 2 . Ans . The characteristic equation is 2 - λ - 3 1 1 - 2 - λ 1 1 - 3 2 - λ = 0 . i.e. - λ ( λ - 1 ) 2 = 0 . So λ = 0 or 1 . Solving ( A - λ I ) x = 0 , we get Eigenvector corresponding to λ = 0 : α ( 1 1 1 ) T , α = 0 . Eigenvector corresponding to λ = 1 : x = α 3 1 0 + β - 1 0 1 and x = 0 .

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Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Eigenvalues & Eigenvectors (Cont’d) Examples Example . Find the eigenvalues & the corresponding eigenvectors of C = 1 2 - 2 1 .
Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Eigenvalues & Eigenvectors (Cont’d) Examples Example . Find the eigenvalues & the corresponding eigenvectors of C = 1 2 - 2 1 .

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Lect26 - Chapter 6 Eigenvalues Eigenvectors Diagonalization...

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