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e403k

# e403k - R 3(see 1d 1b No 4 vectors in R 3 must be LD 1c No...

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MAS 3105 Feb 27, 2003 Quiz 4 Key Prof. S. Hudson 1) [40 points] Set B = { v 1 , v 2 , v 3 , v 4 } = { (1 , 0 , 0) T , (1 , 0 , 1) T , (0 , 1 , 1) T , (0 , 0 , 1) T } . So, B is a set of 4 column vectors in R 3 . Answer, and explain brieﬂy: a) Is B a spanning set of R 3 ? b) Is B linearly independent? c) Let C = { v 1 , v 2 , v 4 } . Is C a basis of R 3 ? d) Let D = { v 1 , v 2 , v 3 } . Show that D a basis of R 3 ? e) Find the transition matrix from D to the standard basis of R 3 . 2) [20 points] Choose ONE of these to prove. a) Show that a nonempty subset of a linearly independent set of vectors { v 1 , v 2 ,... v n } must also be linearly independent. [Use the deﬁnition of l.i.]. b) Show that if { v 1 , v 2 ,... v n } is a basis of V , and v V , then v can be written uniquely as a linear combination of { v 1 , v 2 ,... v n } . Answers: [see also Quiz 4 of Summer 2000]. I gave 4 points for a correct ‘yes’ or ‘no’ and 4 points for the explanation in 1abc. Other explanations are possible: 1a) Yes, because even the ﬁrst 3 vectors span
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Unformatted text preview: R 3 (see 1d). 1b) No, 4 vectors in R 3 must be LD. 1c) No, the set is LD since v 1 + v 4-v 2 = 0 (or explain with a determinant). 1d) Combining the vectors into a matrix A , we see det A =-1 6 = 0, so the columns form a basis of R 3 . 1e) It is the matrix A used in 1d). Do NOT compute its inverse! 2a) This was done in class (and probably also appears in my help pages). A good proof should contain the key equation below (or the equivalent): c 1 v 1 + c 2 v 2 + ... + c j v j + 0 v j + 1 + ... + 0 v n = 2b) This is essentially the ﬁrst half of Thm 3.3.2 (page 151). The main goal is to prove ‘uniquely’, but you should also mention that v is a LC of the other vectors (since they are a basis, they span V ... ). 1...
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