How Can Gram-Schmidt Prove dim(w)=dim(V)-1?

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The discussion centers on proving that the dimension of the subspace w, defined as the set of vectors v in V such that = 0 for all x in V, equals dim(V) - 1. The Gram-Schmidt process is recommended for this proof. The user successfully demonstrates that by selecting a basis for V that includes a non-zero vector x and applying Gram-Schmidt, one loses one dimension, confirming that dim(w) = dim(V) - 1. This approach effectively illustrates the relationship between linear independence and dimensionality in vector spaces.

PREREQUISITES
  • Understanding of linear algebra concepts, particularly vector spaces and subspaces.
  • Familiarity with the Gram-Schmidt process for orthonormalization.
  • Knowledge of inner product spaces and properties of inner products.
  • Ability to work with linear independence and basis vectors in vector spaces.
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  • Study the Gram-Schmidt process in detail to understand its application in vector spaces.
  • Explore the properties of inner product spaces and their implications on dimensionality.
  • Investigate linear independence and its role in determining the basis of a vector space.
  • Practice proving dimensional relationships in various vector spaces using different methods.
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Students and professionals in mathematics, particularly those studying linear algebra, as well as educators looking for effective methods to teach dimensionality concepts in vector spaces.

theFuture
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So I'm to show that the non-zero vector w={v e V|<v,x> = 0} for all x in V that dim(w)=dim(V)-1.

It recommends using the Gram-Schmidt process to prove this but I tried to work it out and I couldn't make any sense of it. Any suggestions on how to start this out?

[edit]: nevermind, I got it. If you were curious, start by saying x is an element of the set S that is linearly independent and spans V. Then do G-S on V and you find that you lose an element of the set, so there you have it.
 
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I think what you mean to say is "prove that the subspace consisting of all vectors v such that <v, x>= 0 (where x is non-zero) has dimension dim(V)- 1."
Select a basis for V containing x and perform a "Gram-Schmidt" starting the process with x as the first vector (so Gram-Schmidt will give a basis containing one basis vector in the same direction as x).
 
That's definitely what I meant, but I copied it verbatim from my crappy book (Messer's Linear Algebra) starting with "show the vector . . ." It's good with proofs, not so good with the math.
 

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