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

• Saph
In summary: So the idea is to assume that 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
Saph

## 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.

Last edited:
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##?

Saph
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!

## 1. What is the definition of a subspace?

A subspace is a subset of a vector space that satisfies the three properties of closure under vector addition, closure under scalar multiplication, and containing the zero vector.

## 2. What does it mean for the union of three subspaces to be a subspace?

It means that the union of three subspaces must also satisfy the three properties of a subspace: closure under vector addition, closure under scalar multiplication, and containing the zero vector.

## 3. How can I prove that the union of three subspaces is a subspace?

To prove that the union of three subspaces is a subspace, you must show that it satisfies the three properties of a subspace: closure under vector addition, closure under scalar multiplication, and containing the zero vector. This can be done by showing that any combination of vectors from the three subspaces also belongs to the union of the three subspaces.

## 4. Can the union of three subspaces be a subspace if the three subspaces do not intersect?

Yes, the union of three subspaces can still be a subspace even if the three subspaces do not intersect. This is because the three subspaces can still satisfy the three properties of a subspace individually, and therefore their union will also satisfy these properties.

## 5. Is the union of three subspaces always a subspace?

No, the union of three subspaces is not always a subspace. In order for the union to be a subspace, it must satisfy the three properties of closure under vector addition, closure under scalar multiplication, and containing the zero vector. If any of these properties is not satisfied, then the union will not be a subspace.

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