Math416_HW7_sol

# Math416_HW7_sol - Math 416 - Abstract Linear Algebra Fall...

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Unformatted text preview: Math 416 - Abstract Linear Algebra Fall 2011, section E1 Homework 7 solutions Section 3.3 3.5. Assume A is nilpotent, i.e. A k = for some k ≥ 1. Then we have det( A k ) = det( ) = 0 (det A ) k = 0 ⇒ det A = 0 . 3.6. Assume A and B are similar, i.e. A = SBS- 1 for some invertible matrix S . Then we have det A = det( SBS- 1 ) = (det S )(det B )(det S- 1 ) = (det S )(det B )(det S )- 1 = det B. 3.7. (2 pts) Assume Q is orthogonal, i.e. Q T Q = I . Then we have det( Q T Q ) = det I = 1 (det Q T )(det Q ) = 1 (det Q )(det Q ) = 1 ⇒ det Q = ± 1 . 1 3.8. (1 pt check) 1 x x 2 1 y y 2 1 z z 2 = 1 x x 2 y- x y 2- x 2 z- x z 2- x 2 = 1 x x 2 y- x ( y- x )( y + x ) z- x ( z- x )( z + x ) = ( y- x ) 1 x x 2 1 y + x z- x ( z- x )( z + x ) = ( y- x )( z- x ) 1 x x 2 0 1 y + x 0 1 z + x = ( y- x )( z- x ) 1 x x 2 0 1 y + x 0 0 z- y = ( y- x )( z- x )( z- y ) . Section 3.4 4.1. a. sign σ = sign(5 , 4 , 1 , 2 , 3) =- sign(1 , 4 , 5 , 2 , 3) = sign(1 , 2 , 5 , 4 , 3) =- sign(1 , 2 , 3 , 4 , 5) =- 1 . b. σ 2 = 1 2 3 4 5 5 4 1 2 3 ◦ 1 2 3 4 5 5 4 1 2 3 = 1 2 3 4 5 3 2 5 4 1 . c. σ- 1 = 1 2 3 4 5 5 4 1 2 3- 1 = 5 4 1 2 3 1 2 3 4 5 = 1 2 3 4 5 3 4 5 2 1 . 2 d. sign( σ- 1 ) = sign(3 , 4 , 5 , 2 , 1) =- sign(1 , 4 , 5 , 2 , 3) = sign(1 , 2 , 5 , 4 , 3) =- sign(1 , 2 , 3 , 4 , 5) =- 1 ....
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## This note was uploaded on 11/07/2011 for the course MATH 416 taught by Professor Staff during the Fall '08 term at University of Illinois, Urbana Champaign.

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Math416_HW7_sol - Math 416 - Abstract Linear Algebra Fall...

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