3.5 - Math 2940 Solutions, Fall 2011 Section 3 . 5 1 )- ~v...

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Unformatted text preview: Math 2940 Solutions, Fall 2011 Section 3 . 5 1 )- ~v 1- ~v 2 + 4 ~v 3- 1 ~v 4 = ~ 0. Solving a~v 1 + b~v 2 + c~v 3 = ~ 0 is the same as solving a + b + c = 0, b + c = 0 and c = 0 which we see has ( a, b, c ) = (0 , , 0) as the only solution. This establishes the independence of { ~v 1 ,~v 2 ,~v 3 } . 2 ) As R 4 is r dimensional, the largest independent set can contain at most 4 vectors. The set { ~v 1 ,~v 2 ,~v 3 } is independent - just forget about the first entry to see this! All of these vectors are in the nullspace of the 4 × 1 rank 1 matrix [1 1 1 1]. It’s null space is necessarily 3 dimensional. So 3 is the largest we can do. 14 ) ~v = . 5( ~v + ~w ) + . 5( ~v- ~w ) and ~w = . 5( ~v + ~w )- . 5( ~v- ~w ). The two pairs { ~v, ~w } and { ~v + ~w,~v- ~w ) } span the same space. 16 ) (a) (1 , 1 , 1 , 1) (b) (1 , , ,- 1), (1 , ,- 1 , 0) and (1 ,- 1 , , 0) works. (c) We need the null space of the matrix 1 1 0 0 1 0 1 1 . Reducing, 1 1 0 0 1 0 1 1 → 1 1 0 0- 1 1 1 → 1 1 0 1- 1- 1...
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This note was uploaded on 01/02/2012 for the course MATH 2940 at Cornell.

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3.5 - Math 2940 Solutions, Fall 2011 Section 3 . 5 1 )- ~v...

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