Unformatted text preview: 3.5 9b] is U1 . Use GE to ﬁnd it. Suggestion: always check your answer to such inverse calculations. U = ± 2 21 1 ² so , U1 = ± 1 / 41 / 2 1 / 4 1 / 2 ² and UU1 = I 3a) This is part of HW 3.6 19c. Explain that each column of AB is a LC of the columns of A so that Col ( AB ) ⊆ Col ( A ). So, dim Col ( AB ) ≤ dim Col A , which implies that rank AB ≤ rank A . 3b) This is HW 4.1 20 and was done in class. 3c) Thm 3.6.6. See the proof in the text. Or the one from the lectures, but be sure to explain each step: dim col A = dim col U b/c same dependency relations = number of leading ones in U since columns with ones form a basis = number of nonzero rows in U there’s a 11 correspondence = dim row U since the nonzero rows are LI = dim row A since row A = row U The ﬁrst step should probably be explained more fully (see lecture notes or page 178 and Ex4 on 179). 1...
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 Spring '09
 JULIANEDWARDS
 Linear Algebra, Matrices, dim Col, row equivalent matrices

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