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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 Leons 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 finitedimensional 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 1 , 2 , . . . , k , then v 1 , v 2 , . . . , v k are linearly independent. Theorem 2 Let 1 , 2 , . . . , 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 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 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 onedimensional. 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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 Spring '07
 Parzen

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