linalg

# linalg - 1 Eigenvalues and Eigenvectors 1.1 Characteristic...

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Unformatted text preview: 1 Eigenvalues and Eigenvectors 1.1 Characteristic Polynomial and Characteristic Equa- tion Procedure. How to find the eigenvalues? A vector x is an e.vector if x is nonzero and satisfies Ax = λx ⇒ ( A- λI ) x = 0 must have nontrivial solutions ⇒ ( A- λI ) is not invertible by the theorem on prop- erties of determinants ⇒ det( A- λI ) = 0 ⇒ Solve det( A- λI ) = 0 for λ to find eigenvalues. Definition . P ( λ ) = det( A- λI ) is called the charac- teristic polynomial . det( A- λI ) = 0 is called a charac- teristic equation . Proposition. A scalar λ is an e.v. of a n × n matrix if λ satisfies P ( λ ) = det( A- λI ) = 0 . Example. Find the e.v. of A = 1- 6 5 ¶ . 1 Since A- λI = 1- 6 5 ¶- λ λ ¶ =- λ 1- 6 5- λ ¶ , we have the characteristic equation det( A- λI ) =- λ (5- λ ) + 6 = ( λ- 2)( λ- 3) = 0 . So λ = 2, λ = 3 are eigenvalues of A . Theorem. Let A be a n × n matrix. Then A is invertible if and only if: a) λ = 0 is not an e.v. of A ; or b) det A 6 = 0. Proof. For b) we have discussed the proof on the de- terminant section. For a): ( ⇒ ): Let A be invertible ⇒ det A 6 = 0 ⇒ det( A- I ) 6 = 0 ⇒ λ = 0 is not an e.v. ( ⇐ ). Let 0 be not an e.v of A ⇒ det( A- I ) 6 = 0 ⇒ det A 6 = 0 ⇒ A is invertible. Theorem. The eigenvalues of a triangular matrix are the entries of the main diagonal. 2 Proof. Recall that a determinant of a triangular ma- trix is a product of main diagonal elements. Hence, if A = a 11 a 12 . . . a 1 n a 22 . . . a 2 n . . . . . . . . . . . . . . . a nn , then the characteristic equation is det( A- λI ) = a 11- λ a 12 . . . a 1 n a 22- λ . . . a 2 n . . . . . . . . . . . . . . . a nn- λ = ( a 11- λ )( a 22- λ ) . . . ( a nn- λ ) = 0 ⇒ a 11 , a 22 , . . . , a nn are the eigenvalues of A . Example. Find the eigenvalues of A = 3 2 3 0 6 10 0 0 2 Solution. det( A- λI ) = det 3- λ 2 3 6- λ 10 2- λ . Thus the characteristic equation is (3- λ )(6- λ )(2- λ ) = ⇒ eigenvalues are 3, 6, 2. Example. Suppose λ is e.v. of A . Determine an e.v. of A 2 and A 3 . What is an e.v. of A n ? 3 Solution. Since λ is e.v. of A ⇒ ∃ nonzero vector x such that Ax = λx ⇒ AAx = Aλx = λAx = λ 2 x . Therefore λ 2 is an e.v. of A 2 . Analogously for A 3 . We have Ax = λx and A 2 x = λ 2 x ⇒ AA 2 x = A 3 x = Aλ 2 x = λ 2 Ax = λ 3 x. Thus λ 3 is an e.v. of A 3 . In general, λ n is an e.v. of A n . 1.2 Similar Matrices Definition. A n × n matrix B is called similar to matrix A if there exists an invertible matrix P such that B = P- 1 AP . Theorem. If n × n-matrices A and B are similar, then they have the same characteristic polynomial and hence the same eigenvalues....
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## This note was uploaded on 02/02/2011 for the course MAE 107 taught by Professor Tsao during the Spring '06 term at UCLA.

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linalg - 1 Eigenvalues and Eigenvectors 1.1 Characteristic...

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