This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: COT 4501 Fall 2010 Homework 3 Solutions 2.25: Let u = ( u 1 , u 2 , . . . u n ) T and v = ( v 1 , v 2 , . . . v n ) T a) uv T = ( v 1 u , v 2 u , . . . , v n u ) T which can be interpreted as each column being a scalar multiple of u . The maximum number of independent columns is one and therefore the rank is one since the vectors u and v are nonzero. b) If rank( A ) = 1 , then the maximum number of independent columns is 1 and A can be written as A = ( a 1 u , a 2 u , . . . , a n u ) where a i is a scalar and u is the basis of the columns. Therefore, A = u ( a 1 , a 2 , . . . , a n ) = uv T , where v = ( a 1 , a 2 , . . . , a n ) T . 2.30: p = 1 : ∥ A ∥ 1 = max j ∑ m i =1  a ij  . Properties 1, 2 and 3 are straightforward to check. Let us now check properties 4 and 5. With C def = A · B , let j be the column index where the maximal sum max j ∑ m i =1  c ij  is attained. Then ∥ AB ∥ 1 = ∥ C ∥ 1 = ∑ i  c ij  = ∑ i  ∑ k a ik b kj  ≤ ∑ i ∑ k  a ik ·...
View
Full Document
 Spring '08
 Davis
 Linear Algebra, Vector Space, Diagonal matrix, Orthogonal matrix, xT Ax

Click to edit the document details