Some Nagging Questions

[kamil9876]

You mean the vectors where and are two different parralel vectors? In that case the set is linearly DEPENDENT. What's the definition of linearly independent in specialist?

[brightsky]

You mean the vectors where and are two different parralel vectors? In that case the set is linearly DEPENDENT. What's the definition of linearly independent in specialist?

a set of vectors {a,b,c} are linearly independent if the only solution to pa + qb + rc = 0 where p, q and r E R is p=q=r=0. i think that's it...but yeah moekemo's spiel has confused me.

[dc302]

another way to say this is that there are no values for m, n such that i=m*j + n*k, hence the vectors are linearly independant.

Is this what confused you? i=m*j + n*k, you can rewrite this as 0 = mj + nk - i and you can see this becomes the same definition as the one you have.

[kamil9876]

You mean the vectors where and are two different parralel vectors? In that case the set is linearly DEPENDENT. What's the definition of linearly independent in specialist?

a set of vectors {a,b,c} are linearly independent if the only solution to pa + qb + rc = 0 where p, q and r E R is p=q=r=0. i think that's it...but yeah moekemo's spiel has confused me.

Right that's the definition* we all like. I think the confusion comes from the phrase "v is linearly dependent on..." which I've never seen, and would be interested to know where it came from  Having said that, whatever that phrase may mean you can always go back to the basic definition to un-confuse yourself.

So using that definition you can immediately see that if is parralel to then for some . So you can see from here that:



Which is a non-trivial solution (ie at least one of the scalars is non-zero, namely the )

Note: It's possible that the set \{a,b,c\} is linearly independent even though none of the vectors are parralel to each other, e.g since and again some of the scalars are non-zero)

*Also should add that are all different, otherwise the set has less than 3 elements and you may get some confusion. (and yeah you can likewise define linear indepence for a set with any number of vectors you like).

note: just used for scalar multiplication just to add clarity in explanation, but may want to avoid that to not confuse with dot product.

[kamil9876]

another way to say this is that there are no values for m, n such that i=m*j + n*k, hence the vectors are linearly independant.

Is this what confused you? i=m*j + n*k, you can rewrite this as 0 = mj + nk - i and you can see this becomes the same definition as the one you have.

In general, if a vector p can be expressed as a linear combination of other vectors q and r (so p=m*q+n*r) then the three vectors are linearly dependent. So to prove this you'd say let p=m*q+n*r then solve for m, n. If there are solutions then the three vectors p,q,r are linearly dependent. If there are no solutions, then the vectors are linearly independent.

I'd like to point out another possible trap here, it is true that if is a linear combination of and then p,q,r is linearly DEPENDENT. But the converse is not true, ie: it is possible that p is NOT a linear combination of q and r, but the vectors p,q,r are still linearly DEPENDENT (trivial counterexample, p=j, q=i, r=2i). However the correct result is that if neither of the three vectors a,b,c can be expressed as linear combinations of the other two, then p,q,r is linearly independent.

[brightsky]

another way to say this is that there are no values for m, n such that i=m*j + n*k, hence the vectors are linearly independant.

Is this what confused you? i=m*j + n*k, you can rewrite this as 0 = mj + nk - i and you can see this becomes the same definition as the one you have.

no her response to this:

just to clarify, if you have 3 vectors a,b,c and a,b are parallel vectors, is it true to say a is linearly dependent on the vectors b,c/b is linearly dependent on the vectors a,c BUT c is linearly independent of the vectors a,b?

the above (and moekamo's answer) sounds logically sound to me, but kamil9876 said that the set of vectors are linearly dependent.

[kamil9876]

Whatever the phrase "linearly dependent/independent on/off" may mean, just go back to the basic definition if confused. I for one have never come across it and secondly it smells very fishy to me since it adds to the many confusions here when trying to relate it to linear independence (ie just because two things sound similair doesn't mean they have to be related).

[dc302]

High school maths is so hard :<

[generalkorn12]

Hi again,
I'm having trouble finding the derivative of arctan ⁴/_x, the answer i'm getting is 4x²/x²+16, yet the book seems to give -4/x²+16.

Is there an explanation for this?

And, for stating the domain of arccos, arcsin and arctan, is there a general equation for determining them?

[b^3]

Let

[kamil9876]

by the chain rule the derivative of it is:

[pi]

I always do it kamil's way, faster than introducing a new variable, remember the shorter way:

In this case, and and then remember your index rules

[kamil9876]

^ lol in the words of a famous professor "the derivative of e to the junk is the derivative of the junk times e to the junk". I think stating the general chain rule like that may be a bit of a mouthful but you get the idea