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 ďHey it is linearly dependentĒ, 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 itís 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)Ö+Km(Am)=0); that there is going to be a value of - Iím 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 Iím 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 Kís 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 Kiís 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 Iíll get K1(A1)+K2(A2)Ö.Kp-1(Ap-1)+Kp+1(Ap+1)Ö.Km(Am)= -Kp(Ap) and I can see that - hey I can derive both sides by ĖKp the reason why I can do that is because I already said hey Kp is not equal to zero so that gives you ĖK1/Kp ĖK2/Kp +Ö. ĖKm/Kp= Ap.
So all Iím doing is dividing by the scalar ĖKp here. And this is what I get and your only finding out that what I basically have is that Iím 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 thatís what we are asked to prove and thatís what we have proved. And thatís the end of this segment