Lect3-07-web

# Lect3-07-web - MATH 304 Fall 2011 Linear Algebra Homework...

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Unformatted text preview: MATH 304, Fall 2011 Linear Algebra Homework assignment #11 (due Wednesday, November 23, 5pm) All problems are from Leon’s book (8th edition). Section 6.1: 1b, 1e, 1f, 1h, 8 Section 6.3: 1b, 1d, 1f, 4b, 6 MATH 304 Linear Algebra Lecture 23: Diagonalization. Review for Test 2. Diagonalization Let L be a linear operator on a finite-dimensional vector space V . Then the following conditions are equivalent: • the matrix of L with respect to some basis is diagonal; • there exists a basis for V formed by eigenvectors of L . The operator L is diagonalizable if it satisfies these conditions. Let A be an n × n matrix. Then the following conditions are equivalent: • A is the matrix of a diagonalizable operator; • A is similar to a diagonal matrix, i.e., it is represented as A = UBU − 1 , where the matrix B is diagonal; • there exists a basis for R n formed by eigenvectors of A . The matrix A is diagonalizable if it satisfies these conditions. Otherwise A is called defective . Theorem 1 If v 1 , v 2 , . . . , v k are eigenvectors of a linear operator L associated with distinct eigenvalues λ 1 , λ 2 , . . . , λ k , then v 1 , v 2 , . . . , v k are linearly independent. Theorem 2 Let λ 1 , λ 2 , . . . , λ k be distinct eigenvalues of a linear operator L . For any 1 ≤ i ≤ k let S i be a basis for the eigenspace associated with the eigenvalue λ i . Then the union S 1 ∪ S 2 ∪···∪ S k is a linearly independent set. Corollary Let A be an n × n matrix such that the characteristic equation det( A − λ I ) = 0 has n distinct real roots. Then (i) there exists a basis for R n consisting of eigenvectors of A ; (ii) all eigenspaces of A are one-dimensional. Example. A = parenleftbigg 2 1 1 2 parenrightbigg . • The matrix A has two eigenvalues: 1 and 3. • The eigenspace of A associated with the eigenvalue 1 is the line spanned by v 1 = ( − 1 , 1). • The eigenspace of A associated with the eigenvalue 3 is the line spanned by v 2 = (1 , 1). • Eigenvectors v 1 and v 2 form a basis for R 2 . Thus the matrix A is diagonalizable. Namely, A = UBU − 1 , where B = parenleftbigg 1 0 0 3 parenrightbigg , U = parenleftbigg − 1 1 1 1 parenrightbigg . Example. A = 1 1 − 1 1 1 1 0 0 2 . • The matrix A has two eigenvalues: 0 and 2. • The eigenspace corresponding to 0 is spanned by v 1 = ( − 1 , 1 , 0). • The eigenspace corresponding to 2 is spanned by v 2 = (1 , 1 , 0) and v 3 = ( − 1 , , 1). • Eigenvectors v 1 , v 2 , v 3 form a basis for R 3 . Thus the matrix A is diagonalizable. Namely, A = UBU − 1 , where B = 0 0 0 0 2 0 0 0 2 , U = − 1 1 − 1 1 1 0 0 1 . Problem. Diagonalize the matrix A = parenleftbigg 4 3 0 1 parenrightbigg ....
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## This note was uploaded on 03/19/2012 for the course STAT 211 taught by Professor Parzen during the Spring '07 term at Texas A&M.

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Lect3-07-web - MATH 304 Fall 2011 Linear Algebra Homework...

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