Prove :Union of Three subspaces is a subspace if ....

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Homework Help Overview

The discussion revolves around proving that the union of three subspaces is a subspace under specific conditions, particularly when one subspace contains the others. The subject area pertains to linear algebra and vector spaces.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants explore the conditions under which the union of subspaces can be considered a subspace, with one participant attempting to prove the case for three subspaces based on a known result for two subspaces. Others suggest examining the implications of the relationships between the subspaces.

Discussion Status

The discussion is ongoing, with participants sharing insights and approaches. One participant has successfully addressed the case for two subspaces but is seeking further guidance on extending this reasoning to three subspaces. There is no explicit consensus yet, but productive lines of inquiry are being explored.

Contextual Notes

One participant notes that they are still learning to format their posts correctly, which may affect the clarity of their contributions. The discussion is focused on theoretical aspects rather than practical examples or applications.

Saph
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Homework Statement


Prove the the union of three subspaces is a subspace if one of the subspaces contains the others

Homework Equations


A subset W of a vector space V is called a subspace if : 1) ##0 \in W ##. 2) if ##U_1## and ##U_2## are in ##W##, then
##U_1 + U_2 \in W##, 3) if ##\alpha ## is a scalar, then ##\alpha U\in W##

The Attempt at a Solution


assume that ##\exists~x,y,z \in U_1\cup U_2\cup U_3 ~## such that, ##x \in U_1 ~, y \in U_2 ~ and~~ z \in U_3##.
We know that, ##x+y+z~\in U_1\cup U_2\cup U_3##, hence ##x+y+z~is~in~either~U_1 ~or~U_2 ~or ~U_3##
Assume, WOLOG, that ##x+y+z~\in~U_1 ,~then~ y+z \in U_1 ,~moreover,~y+z\in U_1 \cup U_2~##,thus
##z\in U_1 \cup U_2~,~and~we~have~two~cases~to~consider##.
##i)~ z \in U_1 ~,~then~y+z\in U_1 ,~\implies~y\in U_1 ~, thus,~ any~z\in~U_3 ~, then~z\in U_1~,~and~any~y \in~U_2 ~, then~y\in U_2##
##hence,~U_2 ~and~U_3 ~\subset U_1##
##ii) ~z\in U_2##, then I don't know how to proceed.
 
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Hi, you can start considering two. Assume that ##U## and ##W## are subspaces of a vector space ##V##. It is possible to prove that if ##U\cup W## is a subspace then either ##U\subseteq W## or ##W\subseteq U##. The idea is that: assume ##U\not\subseteq W## and ##W\not\subseteq U## and pick ##u\in U## and ##w\in W## with ##u\not\in W## and ##w\not\in U## then look at the sum ##u+w\in U\cup W##?
 
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Ssnow said:
Hi, you can start considering two. Assume that ##U## and ##W## are subspaces of a vector space ##V##. It is possible to prove that if ##U\cup W## is a subspace then either ##U\subseteq W## or ##W\subseteq U##. The idea is that: assume ##U\not\subseteq W## and ##W\not\subseteq U## and pick ##u\in U## and ##w\in W## with ##u\not\in W## and ##w\not\in U## then look at the sum ##u+w\in U\cup W##?
Hello, thank you for your answer, I have proved the case for the union of two subspaces (I used the same idea that you suggested ) , my problem is the union of three subspaces.
The post is not complete yet, as I'am learning how to post using latex,I intended to delete this thread but I couldn't, so right now I'am editing the thread to fix latex problems and include my proposed answer.
 
ok!
 

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