# Basis of a Linear Manifold

• Aug 30th 2010, 07:59 AM
Hartlw
Basis of a Linear Manifold
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.
• Aug 30th 2010, 10:03 AM
Hartlw
A Basis
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.
• Sep 3rd 2010, 06:59 AM
Hartlw
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?
• Sep 4th 2010, 10:12 AM
Jose27
Quote:

Originally Posted by Hartlw
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 that \$\displaystyle aX=0\$ then \$\displaystyle a=0\$ or \$\displaystyle X=0\$ (by the properties of being a linear manifold) so any nonzero \$\displaystyle X\$ we have \$\displaystyle \{ X\}\$ is l.i. now take \$\displaystyle S= \{ A\subset M : A\ \mbox{is linearly independent} \}\$ 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).