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Lect27

# Lect27 - Chapter 6 Eigenvalues Eigenvectors Diagonalization...

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Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Diagonalization Motivation Recall our problem again: Find an invertible S such that S - 1 AS = D is a diagonal matrix. Alternative formulation: Let L : n n be defined by L ( x ) = A x . The question is: Can we find a basis F for n such that the matrix representation of L w.r.t. F is D ?

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Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Diagonalization Motivation Recall our problem again: Find an invertible S such that S - 1 AS = D is a diagonal matrix. Alternative formulation: Let L : n n be defined by L ( x ) = A x . The question is: Can we find a basis F for n such that the matrix representation of L w.r.t. F is D ? If we could , what were the vectors in the basis F ?
Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Diagonalization Motivation Recall our problem again: Find an invertible S such that S - 1 AS = D is a diagonal matrix. Alternative formulation: Let L : n n be defined by L ( x ) = A x . The question is: Can we find a basis F for n such that the matrix representation of L w.r.t. F is D ? If we could , what were the vectors in the basis F ? Ans. Eigenvectors.

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Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Diagonalization Motivation Recall our problem again: Find an invertible S such that S - 1 AS = D is a diagonal matrix. Alternative formulation: Let L : n n be defined by L ( x ) = A x . The question is: Can we find a basis F for n such that the matrix representation of L w.r.t. F is D ? If we could , what were the vectors in the basis F ? Ans. Eigenvectors. The question becomes: Can we find a basis consisting of eigenvectors?
Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Diagonalization Theorem 6.3.1 Theorem 6.3.1 Let A be a square matrix of order n . Suppose λ 1 , λ 2 , ··· , λ k are distinct eigenvalues of A , and x 1 , x 2 , ··· , x k are the corresponding eigenvectors. Then x 1 , x 2 , ··· , x k is linearly independent.

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Chapter 6. Eigenvalues & Eigenvectors, Diagonalization Math1111 Diagonalization Theorem 6.3.1 Theorem 6.3.1 Let A be a square matrix of order n . Suppose λ 1 , λ 2 , ··· , λ k are distinct eigenvalues of A , and x 1 , x 2 , ··· , x k are the corresponding eigenvectors. Then x 1 , x 2 , ··· , x k is linearly independent.
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