HW11solutions - Math 221, Solutions for assignment due Nov....

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Unformatted text preview: Math 221, Solutions for assignment due Nov. 11 Section 7.2 Problem 10 det 3- 4- 1- - 2 7 3- = (- 3- )(- 1- )(3- )- 2(- 4(- 1- )) =- ( +1) 2 ( - 1) So the matrix has eigenvalues = 1, with algebraic multiplicity 1 and =- 1, with algebraic multiplicity 2. Problem 12 det 2- - 2 1- 1- 3- - 4 2- 3- = det bracketleftbigg 2- - 2 1- 1- bracketrightbigg det bracketleftbigg 3- - 4 2- 3- bracketrightbigg = ( 2- )( 2- 1) = ( + 1)( - 1) 2 . So The eigenvalues are 0 ,- 1 (with algebraic multiplicity 1) and- 1 (with algebraic multi- plicity 2). Problem 19 This is true. The characteristip polynomial is 2- ( trace ) - det with det < 0, so the discriminant of this polynomial is D = trace 2- 4 det . Since the term under the square root sign is strictly positive, the characteristic polynomial has two distinct roots: trace + D 2 and trace- D 2 . Problem 22 det( A- I ) = det( A- I ) T = det( A T- I ). Thus the characteristic polynomials of A and A T are the same, which implies that the eigenvalues are the same. Problem 22 If B = S- 1 AS , then det( A- I ) = det S- 1 ( A- I ) S = det( S- 1 AS- S- 1 IS ) = det( S- 1 AS- I ) = det( B- I ) ....
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This homework help was uploaded on 02/23/2008 for the course MATH 2210 taught by Professor Pantano during the Fall '05 term at Cornell University (Engineering School).

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HW11solutions - Math 221, Solutions for assignment due Nov....

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