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Given the definition of a linear manifold as a set M of undefined elements "X" and scalars "a" from a real field F; and an association X+Y between elemets of M, and an association aX between elements F and M, such that:
X+Y = Y+X
a(X+Y) = aX + aY
etc
How do you prove the existence of elements of M for which aX + bY + cZ +.. = 0 has no solution other than a = b = c = ... = 0. Or, more basically, elements X and Y such that aX +bY= 0 has no solution other than a = 0 and b= 0.
Consider aX + bY = 0
Either a Y exists st aX + bY = 0 only if b=0, in which case X and Y are Linearly Independent, or Y = pX for all Y and M is at least one-dimensional.
If X and Y are LI, consider aX + bY + cZ = 0.
Either a Z exists st aX + bY +cZ = 0 only if c=0, in which case X, Y and Z are LI, or Z = pX +qY for all Z and M is at least two-dimensional.
Eventually you reach a point where no W exists st aX + bY + cZ +....+ dU + eW = 0 only if e = 0, and then every W can be expressed as a linear combination of X, Y, Z,....,U which is a basis of M of dimension r (number of LI members X,Y,Z....,U).
"Vector" was never mentioned.
I would like to reword original question.
Can you prove from the definition of a linear manifold that linearly independent vectors must exist?
The standard definition of linear independence is: if aX+bY+cZ+..=0 only has a solution for a=b=c..=0, then X,Y,Z.. are linearly independent. Is that an existence proof?
No, that's just a definition. For a proof notice thatQuote:
Originally Posted by Hartlw
then
or
(by the properties of being a linear manifold) so any nonzero
we have
is l.i. now take
and use Zorn's lemma to guarantee the existence of a maximal l.i. set.
The argument you use in your second post works if your manifold is finite dimensional (a fact one does not know before actually finding a basis).