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Unformatted text preview: COT 4501 Fall 2010 Homework 2 Solutions 2.1 We prove equivalence relations for the four definitions of non singularity. 1 ⇒ 2: If ∃ A − 1 such that AA − 1 = I , then det( AA − 1 ) = det A · det A − 1 = det I = 1. Therefore det A ̸ = 0. You need to use the result that det( AB ) = det( A ) det( B ). 2 ⇒ 3: Assume that rank( A ) < n and let the columns of A be the vectors a 1 , a 2 ,..., a n . Since the rank is less than n , we may further assume (since there are only n − 1 independent columns) that a n is a linear combination of a 1 , a 2 , ..., a n − 1 . Then, after first subtracting from the n th column the linear combination, we obtain a new matrix A ′ = ( a 1 | a 2 | ... | a n − 1 | ) which has the same determinant as A since we merely subtracted a linear combination of the first n − 1 columns from the n th column. But, det( A ′ ) = 0 since the n th column is all zero. (This is another property of determinants.) We have obtained a contradiction since we have already shown above that det( A ) ̸ = 0. 3 ⇒ 4: Let z = ( z 1 ,z 2 ,z 3 ,...z n ) T be a column vector. Let A = ( a 1 | a 2 | a 3 | ... | a n ). Since z ̸ = , we can always assume without loss of generality that z n ̸ = 0. If A z = , we have that ∑ k z k z n a k = −...
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This note was uploaded on 01/22/2012 for the course COT 4501 taught by Professor Davis during the Spring '08 term at University of Florida.
- Spring '08