Question

Let v_1, v_2 and v_3 be linearly independent vectors in a vector space V . Prove
that w_1 = 2*v_1, w_2 = 2*v_1 + v_2 + v_3 and w_3 = v_1 + v_2 - v_3 are also linearly independent... Help? Please? :)

Answer 1

suppose it is not L.I. and draw a contadiction

Answer 2

or use the contrapositive

Answer 3

contadicion? lol you seem to be the one to answer these questions of mine every time! i appreciate the patience :P

Answer 4

assume \[c_1w_1+c_2w_2+c_3w_3=0 \]
but at least on of the \[c_i\neq 0\]

Answer 5

expand using the definitions of w1,w2,w3

Answer 6

hopefully this will work...I'm just thinking out loud

Answer 7

\[c_12v_1+c_2(2v_1 + v_2 + v_3)+c_2( v_1 + v_2 - v_3)=0
\]

Answer 8

\[(2c_1+2c_2+c_3)v_1+(c_2+c_3)v_2+(c_2-c_3)v_2=0\]

Answer 9

we have to have \[(2c_1+2c_2+c_3)=(c_2+c_3)=(c_2-c_3)=0\]

since the v's are LI

Answer 10

if we solve this system for the c's

we get all the c's are zero...a contradiction

Answer 11

ok?

Answer 12

Yep - I get all of the maths (thanks!) ... But I might be missing the essential definition of LI .. why is it a contradiction??

Answer 13

we originally assumed at least one of the c's was non-zero...then we get that they are all zero

Answer 14

Why do we make that initial assumption?:/

Answer 15

because I'm trying to prove it by contradiction

the def of LI says that the vectors \[v_1,v_2,\ldots,v_n\] are LI iff

\[c_1v_1+c_2v_2+\ldots+c_nv_n=0\]

has only the trivial solution

Answer 16

He's saying that the set of vectors {v1,...,vn} is linearly independent iff all the coefficients are 0

So assume that the set is NOT independent, then prove that a contradiction arises because of this assumption. So the contradiction would then point you in the opposite direction.

Answer 17

Onto it! Thanks guys!!

Answer 18

The opposite direction being that they are in fact linearly independent