Proof Help: Showing S perp contains V perp when S is a subspace of V

  • Thread starter Thread starter bodensee9
  • Start date Start date
  • Tags Tags
    Proof
Click For Summary
SUMMARY

This discussion focuses on proving that if a subspace S is contained within another subspace V, then the orthogonal complement of S, denoted as S perp, contains the orthogonal complement of V, denoted as V perp. The key argument is based on the relationship between the dimensions of the subspaces and their orthogonal complements, specifically that S + S perp and V + V perp both equal the same dimension N. The proof is established by demonstrating that if a vector x is in V perp, it must also be in S perp due to the subset relationship between S and V.

PREREQUISITES
  • Understanding of vector spaces and subspaces
  • Knowledge of orthogonal complements in linear algebra
  • Familiarity with inner product notation and properties
  • Basic concepts of dimension in vector spaces
NEXT STEPS
  • Study the properties of orthogonal complements in linear algebra
  • Learn about the dimension theorem for vector spaces
  • Explore examples of subspaces and their orthogonal complements
  • Investigate the implications of the rank-nullity theorem
USEFUL FOR

Students and educators in linear algebra, mathematicians focusing on vector space theory, and anyone seeking to understand the relationships between subspaces and their orthogonal complements.

bodensee9
Messages
166
Reaction score
0

Homework Statement



Can someone help me with this proof? I'm supposed to show that if a subspace S is contained in subspace V, then S perp contains V perp.

Homework Equations



None, or the dimensions must add up, so S + S perp = some dimension N, and V + V perp equals the same dimension N, if S and V are both subspaces of N.

The Attempt at a Solution



Am I supposed to show that the dimensions don't add up? Can anyone provide suggestions? Thanks.
 
Physics news on Phys.org
S perp contains V perp iff x in V perp implies x in S perp.
x in V perp iff for all y in V, <x,y> = 0.
x in S perp iff for all y in S, <x,y> = 0.

So you want to show (for all y in V, <x,y> = 0) implies (for all y in S, <x,y> = 0). Well S is a subset of V, so this is obvious.
 

Similar threads

Replies
34
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
Replies
2
Views
10K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 15 ·
Replies
15
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 5 ·
Replies
5
Views
7K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
4K