11 - Eigen 3 Diagonalization

# 11 - Eigen 3 Diagonalization - 1 Math 1b Practical...

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1 Math 1b Practical — Eigenvalues and eigenvectors, III: Diagonalization February 14, 2011 If we can express a vector x as a linear combination of eigenvectors of an n × n matrix A , then we may compute (±nd a formula for) A n x and so understand what happens when n is large. To this end, it is good thing when we can ±nd a basis for R n consisting of eigenvectors of A . Important: Note that there are matrices for which no basis of eigenvectors exists. See the examples below Theorem 3. (There are things one can do to understand A n x even when x is not in the span of the eigenvectors of A , but we will not have time to get into this in detail in this course.) Theorem 1. If e 1 , e 2 ,..., e k are eigenvectors and λ 1 2 ,...,λ k the corresponding eigen- values of a matrix A , and if the eigenvalues λ i are distinct, then the eigenvectors e i are linearly independent. Proof: We proceed by induction on k . The assertion is true for k = 1 because we have required that eigenvectors be nonzero vectors. Assume the assertion holds for k = ` .W e will show it holds for k = ` +1. Suppose c 1 e 1 + c 2 e 2 + ... + c ` e ` + c ` +1 e ` +1 = 0 . (2) Multiplying by A on the left, we get c 1 A e 1 + c 2 A e 2 + + c ` A e ` + c ` +1 A e ` +1 = 0 , or c 1 λ 1 e 1 + c 2 λ 2 e 2 + + c ` λ ` e ` + c ` +1 λ ` +1 e ` +1 = 0 . (3) Multiply (2) by λ ` +1 and subtract it from (3) to get c 1 ( λ 1 λ ` +1 ) e 1 + c 2 ( λ 2 λ ` +1 ) e 2 + + c ` ( λ ` λ ` +1 ) e ` = 0 .

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11 - Eigen 3 Diagonalization - 1 Math 1b Practical...

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