linear dependence alternate method?

[almostatrap]

So the standard definition is for linear dependence of vectors a, b and c:     ka + lb + mc = 0 , where not all of k, l and m are 0

What I have in my notes from my teacher is   ka + lb = c

This seems to work (and is easier i guess), but im not sure. Can someone please clarify. Is this definition sufficient?

Thanks

[TrueTears]

It's the same.

m just equals 1 in your teacher's notes.

[almostatrap]

cool. Is it okay on an exam for a "prove that... are linearly dependent" question?

[TrueTears]

cool. Is it okay on an exam for a "prove that... are linearly dependent" question?

It's fine.

[dcc]

I'd be careful when considering sets of vectors which contain the zero-vector and using this method.  I suspect something fishy might happen, but I can't think of any particular example.  In that case, just remembe that any set of vectors which contains the zero vector is linearly dependent.

[kamil9876]

I think a problematic situation arises when one of the scalars, namely is zero.

You can derive one expression from the other by some manipulation:





but the two scalars here are real numbers hence the alternative expression(i.e: just let and . However if it is required that m=0, then the above argument fails and the failure can be demonstrated by this example:

if then a solution to the first equation is Hence the set of vectors is linearly dependent since not all the scalars are zero. However the second equation gives:




which has no solutions for and hence you may wrongly conclude that the set is linearly independant. The reason why this does not work is because you implicitly divided by 0. However if you did find a solution to the second equation then there would be a solution to the first hence this method is better when wanting to prove linear dependance, but not very good at proving linear independance since you would have to take every labeling of vectors into acount (i.e if you relabel the same set of vectors then you find that there is a solution ()

my tip: go with the first for proving that it's linearly independant, but if wanting to prove linear dependance and you see a more obvious solution (which is ussually the case) then go for it.

[ice_blockie]

what kamil said. My teacher EXPLICITLY told us to use the first formula to prove when something is independent or dependent and only use the second one when the question says something along the lines of "prove that...is linearly dependent" etc.

[kamil9876]

yep. the first formula takes everything into account. The second is only a special case. In order to prove dependance finding just one example of a linear relationship between the vectors suffices (and the second formula can do this). Proving independance is more tricky since you must cover ALL possiblities, and the second formula does not do that (such as the case of m=0 is not taken into account)

[mano91]

can anyone do a question from essentials using both methods as an example?

show that vectors  a=8i + 5j + 2k, b= 4i -3j + k  and c=2i -j +0.5k are linearly dependent.

i always use ma +nb =c  and it works for me :S but the other method confuses me.

and another one, a=8i + 5j + 2k, b= 4i -3j + k  and c=2i -j +0.5k are linearly independent.

thank you

[NE2000]

yep. the first formula takes everything into account. The second is only a special case. In order to prove dependance finding just one example of a linear relationship between the vectors suffices (and the second formula can do this). Proving independance is more tricky since you must cover ALL possiblities, and the second formula does not do that (such as the case of m=0 is not taken into account)

Most of the questions I've seen on exams say its a non-zero vector. Are there any other exceptions?

[kamil9876]

NE2000: is a zero scalar in my example. I can't think of an example where the zero vector is as problematic, however the special thing about this is that any set of vectors that contains the zero vector is automatically a dependant set. (prove this using the "second formula").

Mano91: for the first question ma+bn=c is fine for the first question( i can imagine how ridiculous and unnecesarily over-complicated it would be using the the other method).

Second question: by using ma+nb+lc=0 you get three equations(from equating each component to 0). As an example, the one from equating i component to 0 would be:

8m+4n+2l=0

If m=n=l=0 is the only solution, then linearly independant.

[kamil9876]

I can't think of an example where the zero vector is as problematic

just thought of one. c=0, a=i, b=j.

Linearly dependant set yet using ma+nb=c wrongly infers otherwise.

However it's still true that by sticking to:

my tip: go with the first[ma+nb+lc=0] for proving that it's linearly independant, but if wanting to prove linear dependance and you see a more obvious solution (which is ussually the case) then go for it[ma+bn=c].

you will be fine in all cases.

[mano91]

thanks kamil but with the second one u utilised ma + nb+ kc = 0. can it be proven with the other formula?

[kamil9876]

That would be a flaw when proving linear independance..

Take the example i provided earlier. Determine whether the set of vectors {2i,i,3j} is linearly independant. note that

3j=m(2i)+n(i) has no solution. (1)

however:

2i=m(3j)+n(i) has a solution. (m=0, n=2). (2)


(2) shows the set is linearly dependant. (1) would suggest it is linearly independant, but it is not-it is dependant as shown by (2). This is an example of how this method cannot be use to prove independance, because you don't know if by "swapping" two vectors (in our case we swapped the 2i with 3j to get (2)) you will indeed get a linear combination(which would prove linear dependance). Therefore this method is only good at showing dependance. I explained the deeper reason why in the first post.

Anyways, in my opinion showing that three simultaenous equations only have the solution m=0,n=0,k=0 is pretty rare in VCE, not to mention tedious and might even be considered beyond the course.