lect5 - EL 625 Lecture 5 1 EL 625 Lecture 5 Cayley-Hamilton...

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EL 625 Lecture 5 1 EL 625 Lecture 5 Cayley-Hamilton Theorem : Every square matrix satisfies its own characteristic equation If the characteristic equation is p ( λ ) = det( λI - A ) = λ n + a n - 1 λ n - 1 + a n - 2 λ n - 2 + ... + a 1 λ + a 0 From Cayley-Hamilton theorem, p ( A ) = A n + a n - 1 A n - 1 + a n - 2 A n - 2 + ... + a 1 A + a 0 I = Θ Thus, A n can be expressed as a linear combination of the matrices, A i ,i = 0 , 1 , 2 ,...,n - 1 ( A 0 = I ). A n = - n - 1 X i =0 a i A i A n +1 = - n - 1 X i =0 a i A i +1 = - n - 1 X i =0 b i A i A n +1 can be expressed as a linear combination of A,A 2 ,...,A n . But, since A n itself can be expressed as a linear combination of
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EL 625 Lecture 5 2 I,A,A 2 ,...,A n - 1 = A n +1 can be expressed as a linear combi- nation of I,A,A 2 ,...,A n - 1 . Any power of A can be written as a linear combination of I,A,A 2 ,...,A n - 1 Any function which has a convergent power series expansion can be expressed as a linear combination of I,A,A 2 ,...,A n - 1 . f ( A ) = n - 1 X j =0 γ j A j If the matrix, A can be diagonalized via a similarity transformation, T , i.e. T - 1 AT = Λ, T - 1 f ( A ) T = n - 1 X j =0 γ j T - 1 A j T = n - 1 X j =0 γ j ( T - 1 AT ) j = f (Λ) = n - 1 X j =0 γ j Λ j f (Λ) = f ( λ 1 ) 0 ... 0 0 f ( λ 2 ) ... 0 . . . . . . . . . . . . 0 0 ... f ( λ n )
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EL 625 Lecture 5 3 = n - 1 X j =0 γ j Λ j = n - 1 j =0 γ j λ j 1 0 ... 0 0 n - 1 j =0 γ j λ j 2 ... 0 . . . . . . . . . . . . 0 0 ... n
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This note was uploaded on 12/06/2010 for the course EL 625 taught by Professor Khorami during the Spring '10 term at NYU Poly.

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lect5 - EL 625 Lecture 5 1 EL 625 Lecture 5 Cayley-Hamilton...

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