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Tricky subspace & intersection Problem

  • #1

Homework Statement



I am trying to solve this problem:
Let W_1, W_2, W_3 be subspaces of a vector space, V.
Prove that W_1 ∩ (W_2 + ( W_1 ∩ W_3)) = (W_1 ∩ W_2) + (W_1 ∩ W_3).
Can someone help me show this? I have tried using Dedekind's law, but not sure it that is the way to go.


The attempt at a solution

I tried with in my mind very trivial case...can somebody please show me a more detailed solution with more steps?

This is what I did....Since a subspace is a set, the laws of set operations apply. I assume (not sure if this is a valid assumption) that + here is the same as "union".

Now intersection is distributive over union,
i.e. a∩(b+c) = a∩b + a∩c
so in this case,
W_1 ∩ (W_2 + ( W_1 ∩ W_3))
= (W_1 ∩ W_2) + (W_1 ∩ (W_1 ∩ W_3))
In the second term, I use the properties that intersection is associative, and W_1 ∩ W_1 = W_1, and that term becomes W_1 ∩ W_3 which proves the required result.


Now can anyone answer if this always holds and please show me a more detailed solution with more steps that would make more sense? Thanks...
 

Answers and Replies

  • #2
George Jones
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I haven't thought about how to do the question, but
I assume (not sure if this is a valid assumption) that + here is the same as "union".
is not a valid assumption. Have you looked in your notes or text for a definition?
 
  • #3
I haven't thought about how to do the question, but


is not a valid assumption. Have you looked in your notes or text for a definition?
Yes I did...but unfortunately, I am totally lost and confused now...
 
  • #4
George Jones
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Yes I did...but unfortunately, I am totally lost and confused now...
Write down Dedekind's law (in a post).
 
  • #5
Write down Dedekind's law (in a post).
V ∩ (U + W)= U + (V ∩ W) this is what I got....
 
  • #6
George Jones
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V ∩ (U + W)= U + (V ∩ W) this is what I got....
Is there a condition that U and V are required to satisfy?
 
  • #7
U, V and W, are just subsets of a set S, right?
 
  • #9
In my question those are all the conditions specified. Do you mean that U <= V? How does, this help me understand whether the LHS is equal to the RHS? The real question is thus, when is this valid, and how I can get the RHS from LHS?
 
  • #10
George Jones
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The LHS of your question and the LHS of Dedekind's law look similar.

[edit}

To make them the same, take [itex]V = W_1[/itex], and either [itex]U = W_2[/itex] and [itex]W = W_1 \cap W_3[/itex], or [itex]W = W_2[/itex] and [itex]U = W_1 \cap W_3[/itex].

[/edit]

With respect to Dedekind's law, does either of these choices allow you to show the desired result?
 
  • #11
I dont think so:
The first one results in:
RHS = W_1 + (W_2 + W_1 ∩ W_3)

The second one results in:
W_1 + (W_1 ∩ W_3 ∩ W_2)

What should I do now?? What conclusion can I take away from this?
 
  • #12
George Jones
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Take a look at the second choice more carefully.
 
  • #13
I realize that for the equality to hold then:

W_1 + (W_1 ∩ W_3 ∩ W_2) = W_1 ∩ (W_2 + ( W_1 ∩ W_3))

However I can not see how... Could you please guide?
 
  • #14
George Jones
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I'm a little confused.

With the second of my choices in post #10 (note the edit):

what are V, U , and W;

what is the LHS of Dedekind's laws;

what is the RHS of Dedekind's laws?
 
  • #15
I just noticed the edit:
Using the edited U, V, W from post 10 I get that:
W_1 ∩ (W_1 ∩ W_3 +W_2) = (W_1 ∩ W_3) + (W_1 ∩ W_2)

Hence, I have shown the equality in the original problem statement (Thank you).

One question remains and that is... Is this equality ALWAYS valid? Are there any restrictions? I thought the condition in the original Dedekind's law with U <= V could be one. Are there others?
 
  • #16
Dear George,
Could you please help me out regarding my previous questions (In Post #15)? Thank you....
 

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