Could somebody explain linear independence to me? I was asked earlier to show whether not the solutions to some (arbitrary) DE are x, x+1, and x^2, and I wasn't sure what to do or how to approach it.

Question

perl · Answer

would need more information to help you

zzr0ck3r · Answer

A set of elements of some vector space are linearly independent if no one of the elements can be written as a linear combination of the other elements.

example:

Consider the subset of $\mathbb{R}^3$ $\{(1,2,3),(4,5,6),(9,12,15)\} $.
Lets name these vectors $a,b$ and $c$ respectively.

This set is NOT linearly independent because $a+2b=c$.

We might as well remove $c$, because we can "get to it" through the other two vectors.

then $\{a,b\}$ is linearly independent. because $1*4 = 4$ but $2*4 \ne 5$.

hope this helps.

mendicant_bias · Answer

So, none of these are linearly *dependent*, because any constant multiple will not achieve one equalling the other, correct?

mendicant_bias · Answer

e.g. they are all linearly dependent?

mendicant_bias · Answer

(The x, x+1, and x^2)

zzr0ck3r · Answer

so you are in R[x]?

mendicant_bias · Answer

I believe so, but don't really know, I haven't taken Linear Algebra-which sounds relevant-and I'm dealing with sets of solutions to ODE's, e.g. all three of those are proposed as solutions, but whether they are linearly independent or not is what's in question. I don't know if them being ODE's affects the idea of linear independence or not.

zzr0ck3r · Answer

wih polys of max degree 2?

zzr0ck3r · Answer

they are, you cant make any one of these polynomials by adding any of the other polynomials together and multiplying by constants.

mendicant_bias · Answer

Alright; how does the polynomial degree affect the situation?

zzr0ck3r · Answer

nothing really, I just wanted to make sure we were talking about the same thing...

mendicant_bias · Answer

Lol, yeah, we are. Thanks so much.