Linear Algebra dimensions proof

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SUMMARY

The discussion focuses on determining necessary and sufficient conditions for the equality dim(W1 ∩ W2) = dim(W1) in the context of finite-dimensional vector spaces. It establishes that if W1 is a subspace of W2 (W1 ⊆ W2), then the dimension of their intersection equals the dimension of W1. The Replacement Theorem is referenced as a foundational concept in this proof. The participants clarify the implications of the dimensions involved, specifically noting that if dim(W1) = m and dim(W2) = n with n ≥ m, the nesting condition holds.

PREREQUISITES
  • Understanding of finite-dimensional vector spaces
  • Familiarity with the concept of subspaces
  • Knowledge of the Replacement Theorem
  • Basic linear algebra terminology and notation
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  • Study the implications of the Replacement Theorem in linear algebra
  • Explore the properties of subspaces in vector spaces
  • Learn about dimension theory in linear algebra
  • Investigate proofs related to intersections of vector subspaces
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Students and educators in linear algebra, mathematicians focusing on vector spaces, and anyone seeking to deepen their understanding of dimension theory and subspace relationships.

zcd
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Homework Statement


Let W1 and W2 be subspaces of a finite-dimensional vector space V. Determine necessary and sufficient conditions on W1 and W2 so that dim(W_1 \cap W_2)=dim(W_1)

Homework Equations


Replacement Theorem

The Attempt at a Solution


To clarify on the question: is the problem asking for conditions such that condition\iff dim(W_1 \cap W_2)=dim(W_1)?
If it is, is it possible to say that let dim(W1)=m and dim(W2)=n, n≥m and W1 is nested inside W2, so W_1 \subseteq W_2?
 
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zcd said:
To clarify on the question: is the problem asking for conditions such that condition\iff dim(W_1 \cap W_2)=dim(W_1)?

Yes.

zcd said:
If it is, is it possible to say that let dim(W1)=m and dim(W2)=n, n≥m and W1 is nested inside W2, so W_1 \subseteq W_2?
"W_1 \subseteq W_2" is a reasonable guess for what the right condition might be, but what you have written is not a proof of either direction of the implication.
 

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