Re: Dimension Question



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

.



Relevant Pages

  • Re: Dimension Question
    ... perpendicular subspace to U, i.e. any vector from V is perpendicular ... to all vectors from U. Proove that dimU+dimV=dimQ. ... what I believe is the more precise terminology: ...
    (sci.math)
  • Dimension Question
    ... 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. ... Narek Saribekyan ...
    (sci.math)
  • Re: Dimension Question
    ... On Aug 15, 5:04 pm, Narek Saribekyan ... perpendicular subspace to U, i.e. any vector from V is perpendicular ... every function defines a single subset of edges. ...
    (sci.math)