English Deutsch Français Italiano Español Português 繁體中文 Bahasa Indonesia Tiếng Việt ภาษาไทย
All categories

Are the folowing list of vectors linearly independent??

A) { ( 2 , 1 ) , ( 1 , 2 ) , ( 2 , 2 ) }
B) { ( 1 , 0 , -4 , 7 ) , ( 5 , 3 , 1 , 6 ) , ( -2 , -1 , 0 , 8 ) , (0,0,0,0) }
C) { ( 1 , 2 , 4 ) , ( 0 , -1 , 1 ) , ( 3 , -1 , 1 ) }
D) { ( 2 , -1 , 1 ) , ( 4 , 0 , 2 ) }
E) { e1 , e1+e2,......,e1+e2+....+en } where ei in R^n is the vector which has 1 in the i-th place and 0 everywhere else!

HELP!!!

2007-05-07 21:19:34 · 3 answers · asked by Fred B 1 in Science & Mathematics Mathematics

3 answers

A) No; we have two dimensions and three vectors.
B) No; contains the zero vector.
C) Check: a(1, 2, 4) + b(0, -1, -1) + c(3, -1, -1)
= (a + 3c, 2a - b - c, 4a - b - c)
= 0 <=> a = -3c, b+c = 2a, b+c = 4a
=> 2a = 4a and hence a = 0; from a = -3c we get c = 0, and then b = 0 also. So they are linearly independent.
D) Yes, two vectors that are not scalar multiples of each other are linearly independent.
E) Yes. Suppose c1 e1 + c2 (e1+e2) + ... + cn (e1+e2+...+en) = 0. Then we have
(c1+c2+...+cn, c2+...+cn, ..., cn-1 + cn, cn) = 0.
So cn = 0 and by successive back-substitution we get
cn-1 = 0, cn-2 = 0, ..., c2 = 0, c1 = 0.
So they are linearly independent.

2007-05-07 21:49:46 · answer #1 · answered by Scarlet Manuka 7 · 0 0

D and e are right

2007-05-07 22:01:09 · answer #2 · answered by 77e6 e 1 · 0 0

ir2

2007-05-07 21:28:07 · answer #3 · answered by stratoframe 5 · 0 1

fedest.com, questions and answers