5.1(10,11,14,18,20) 5.2(8,11,12)

5.1(10,11,14,18,20) 5.2(8,11,12) - Homework 10 - Solutions...

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Homework 10 - Solutions 5.1 Problem 10 If A is upper-triangular, then so is A λI , and the determinant of a such a matrix is the product of the diagonal entries. 5.1 Problem 11 (a) Since A is similar to a scalar matrix λ I, there exists an invertible matrix S, such that A = S - 1 ( λI ) S = λS - 1 IS = λS - 1 S = λI (b) Since A is a diagonalizable matrix with only one eigenvalue λ , there exists an invertible matrix Q, such that A = Q - 1 ( λI ) Q Using (a), the proof is complete. 5.1 Problem 14 Use the fact that det( B )=det( B t ) for any square matrix B . Apply to the case B = A λI 5.1 Problem 18 (a) Since B is invertible, det( B - 1 ) 6 =0,and det( A + cB )= det( B - 1 ( A + cB )) det( B - 1 ) Let ˜ A = B - 1 A ,c=- λ ,thendet( A + cB )=0iF det( B - 1 A + cI ˜ A λI )=0 This is an equation over the ±eld C , hence it always has at least one root, i.e there exists a c =( 1) λ C such that det( B - 1 A + cI ˜ A λI This implies that det( A + cB ) = 0 and hence A + cB is not invertible. g 1
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5.1 Problem 20 f (0) = det( A 0 I )=det( A ) A is invertible if and only if det( A )= a 0 6 =0, 5.2 Problem 8 Let m 1
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This note was uploaded on 02/02/2012 for the course MATH 115A 262398211 taught by Professor Fuckhead during the Spring '10 term at UCLA.

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5.1(10,11,14,18,20) 5.2(8,11,12) - Homework 10 - Solutions...

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