Proof dim(U+V)=dim U+dim V - dim(U∩V)

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    Dimension Proof
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Discussion Overview

The discussion revolves around proving the relationship between the dimensions of the sum of two subspaces \( U \) and \( V \) of \( \mathbb{R}^n \), specifically the equation \( \text{dim}(U+V) = \text{dim}(U) + \text{dim}(V) - \text{dim}(U \cap V) \). The scope includes theoretical exploration and mathematical reasoning related to vector spaces and their dimensions.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Exploratory

Main Points Raised

  • One participant defines the dimension of a vector space as the number of vectors in any basis for a subspace.
  • Another participant suggests considering the case where \( U \cap V = \{0\} \) to simplify the proof.
  • It is noted that if \( U \cap V = \{0\} \), the intersection of any bases for \( U \) and \( V \) will be the empty set, leading to \( \text{dim}(U \cap V) = 0 \).
  • Participants discuss the importance of finding bases for the vector spaces \( U + V \), \( U \), \( V \), and \( U \cap V \) that are related to each other.
  • One participant emphasizes starting with the basis for \( U \cap V \) to construct bases for \( U \) and \( V \).
  • Another participant suggests counting the vectors in the bases of \( U \), \( V \), and their intersection to establish the relationship.

Areas of Agreement / Disagreement

Participants express various approaches to the proof, with some suggesting specific cases or methods while others propose more general strategies. No consensus is reached on a single method for proving the identity.

Contextual Notes

The discussion includes assumptions about the nature of the bases and their relationships, which may not be universally applicable without further clarification. The steps for generalizing the proof when \( U \cap V \) is not the null subspace remain unresolved.

b00tofuu
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let U and V be subspaces of Rn. Prove that dim(U+V)=dim U+dim V - dim(U∩V)
 
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Okay, first, what is your definition of "dimension" of a vector space?
 
its the number of vector in any basis for a subspace
 
Here's a hint: if [tex]U \cap V = \left\{0\right\}[/tex], then what must be the intersection of any base of U with any base of V?
 
a point...?
but i still don't understand how to write the proof... T_T
 
the zero point
 
Notice that the basis are SETS of vectors: if [tex]U \cap V = \left\{0\right\}[/tex], then the intersection of any base of U, with any base of V will be the empty set; also, [tex]dim\left(U\capV\right) = dim\left(\left\{0\right\}\right) = 0[/tex]. Try this particular case first, then see if can generalize when [tex]U \cap V[/tex] is not the null subspace.
 
One way to approach the problem is to ask yourself if you can find bases for the vector spaces U + V, U, V, and U ∩ V that are related somehow to each other.
 
You are complicating too much; it's simpler than that: consider a basis [tex]\left\{b_{i}\right\}[/tex] for [tex]U \cap V[/tex]; this basis can be extended to a basis [tex]\left\{u_{i}\right\}[/tex] of U and [tex]\left\{v_{i}\right\}[/tex] of V; now it's only a matter of counting the vectors.
 
  • #10
I believe we're talking about the same thing. Extending a basis is what I meant by finding bases for U, V, and U ∩ V that are related to each other, then picking the right vectors from those bases to be a basis for U + V. Of course, I still might be missing something even simpler; it wouldn't be the first time :-)
 
  • #11
Yes, but for proving that identity, not all all basis will do. Start with [tex]U \cap V[/tex].
 
  • #12
Choose a basis for U. If any of those basis vectors are also in V, you can construct a basis for V including those vectors. If not, just choose any basis for V. Of course, the vectors in both bases, if any, form a basis for [itex]U\cap V[/itex]. Now, just count!

How many vectors are there in the basis for U? How many vectors are there in the basis for V? How many are in both?
 

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