This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full Document
 '05
 HUI
 Math, Linear Algebra, Algebra, basis

Click to edit the document details