CHAPTER 02.22: VECTORS: Prove
that if a set of vectors is linearly dependent, then at least one vector can
be written as a linear combination of others In
this segment we will talk about if somebody gives you a set of vectors and
they say that “Hey it is linearly dependent”, then we are asked to show that
at least one vector can be written as a linear combination of others. So the proof
is as follows: Let A1, A2 all the way up to Am be linearly dependent. So it’s
given to us that we have a set of vectors which is linearly dependent; then
we know that if I write down a linear combination such as this one
K1(A1)+K2(A2) all the way up to +Km(Am) and I put that equal to 0 vector (K1(A1)+K2(A2)…+Km(Am)=0);
that there is going to be a value of - I’m going to get a non-previous
solution because its already given to me that A1 through Am is a linearly
dependent set of vectors. So
I’m going to find out the non-previous solution so at least what I want to
find out at least one Ki is not equal to zero so at least one of the K’s has
to not be equal to zero if I want to solve that particular set of equations
right there. So let that be so if one of the Ki’s is not equal to zero let
that be Kp so let the K which is Ki which is not
equal to zero I go from 1 to m let that be Kp where
p is some number which is between 1 and m so if it is the kpth
vector corresponding to which you get a Kp which is
not equal to zero. Then what is the case we have K1(A1)+K2(A2) all the way up
to +Kp(Ap)+ all the way
up to +Km(Am)=0 vector and since we have now said that hey Kp is not equal to zero let me take that to the left hand
side or to the right hand side. In this case I’ll get
K1(A1)+K2(A2)….Kp-1(Ap-1)+Kp+1(Ap+1)….Km(Am)= -Kp(Ap) and I can see that - hey I can derive both sides by –Kp the reason why I can do that is because I already said
hey Kp is not equal to zero so that gives you –K1/Kp –K2/Kp +…. –Km/Kp= Ap. So
all I’m doing is dividing by the scalar –Kp here. And
this is what I get and your only finding out that what I basically have is
that I’m able to write down the the Ap vector as a linear combination of A1, A2, A3 so on and
so forth up to Am. And that’s what we are asked to prove and that’s what we
have proved. And that’s the end of this segment |