Re: Dimension Question
- From: beeworks@xxxxxxxxxxx
- Date: Wed, 15 Aug 2007 07:34:54 -0700
On Aug 15, 8:52 am, Narek Saribekyan <narek.saribek...@xxxxxxxxx>
wrote:
On Aug 15, 5:04 pm, Narek Saribekyan <narek.saribek...@xxxxxxxxx>
wrote:
Hi all, I need help with this.
Let Q be a vertex space; let U be a subspace of Q and V be a
perpendicular subspace to U, i.e. any vector from V is perpendicular
to all vectors from U. Proove that dimU+dimV=dimQ.
Thanks,
Narek Saribekyan
I don't know it make's sense or not, but elements of Q are subsets of
graph edges (or vertices: everyting's the same for both edges and
vertices, so I'll say 'edges'). The field of coefficients is F2 = {0,
1}, (so 1+1=0), and the elements of Q are all functions phi:E->F2, so
every function defines a single subset of edges.
<F,F1>:=dot product
V = {D from Q | <D,F>=0 (perpendicular) F from U), so V is
'perpendicular' to U.
-->From Reinhard Diestel - Graph Theory ||, Section 1.9 - Some linear
algebra.
Your terminology is a little off. Here is your post reworded with
what I believe is the more precise terminology:
Let Q be a _vector_ space; let U be a subspace of Q and V be an
_orthogonal_ subspace to U, i.e. any vector from V is _orthogonal_
to all vectors from U. Proove that dimU+dimV=dimQ.
Counter example:
Q is R3. dim(R3) = 3. U is the x axis, V is the y axis. dim(U) = 1,
dim(V) = 1. dim(U) + dim(V) = 2 /= dim(R3) = 3.
- MO
.
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