# Formulation of a proof of subspaces

• GlassBones
In summary: No, I don't think so.I'm stuck here. I'm thinking to do proof by contradiction? y + W is a subspace and y \not\in W. I'm thinking to find a property it violates? But I don't know how to do this. Any hints on how I should proceed?In summary, if y is in W then any vector in x will satisfy x=y+w such that y,w is in W, so y + W is a subspace of V.
GlassBones

## Homework Statement

Let W be a subspace of a vector space V, let y be in V and define the set $y + W = \{x \in V | x = y +w, \text{for some } w \in W\}$ Show that $y + W$ is a subspace of V iff $y \in W$.

## The Attempt at a Solution

Let W be a subspace of a vector space V, let y be in V and define the set $y + W = \{x \in V | x = y +w,\text{for some } w \in W\}$.
proof(←)
if $y \in W$ then any vector in $x \in y + W$ will satisfy $x=y+w$ such that $y,w \in W$. Since W is a subspace and is closed under addition, all the vectors in $y+W$ must also be in W, i.e. $y + W = W$.

proof(→)
...I'm stuck here. I'm thinking to do proof by contradiction? $y + W$ is a subspace and $y \not\in W$. I'm thinking to find a property it violates? But I don't know how to do this. Any hints on how I should proceed?

GlassBones said:

## Homework Statement

Let W be a subspace of a vector space V, let y be in V and define the set $y + W = \{x \in V | x = y +w, \text{for some } w \in W\}$ Show that $y + W$ is a subspace of V iff $y \in W$.

## The Attempt at a Solution

Let W be a subspace of a vector space V, let y be in V and define the set $y + W = \{x \in V | x = y +w,\text{for some } w \in W\}$.
proof(←)
if $y \in W$ then any vector in $x \in y + W$ will satisfy $x=y+w$ such that $y,w \in W$. Since W is a subspace and is closed under addition, all the vectors in $y+W$ must also be in W, i.e. $y + W = W$.

proof(→)
...I'm stuck here. I'm thinking to do proof by contradiction? $y + W$ is a subspace and $y \not\in W$. I'm thinking to find a property it violates? But I don't know how to do this. Any hints on how I should proceed?
If ##y\notin W##, how would you construct ##0 \in y+W\,?##

I can make 0 by making y + w to both be 0.
Since w = 0 and w is in W, y cannot be 0. So I cannot construct 0 without y in W.

I can also make zero by using the additive inverse of y to be -w. But that would mean y should be in W.

Is this the correct reasoning?

GlassBones said:

## Homework Statement

Let W be a subspace of a vector space V, let y be in V and define the set $y + W = \{x \in V | x = y +w, \text{for some } w \in W\}$ Show that $y + W$ is a subspace of V iff $y \in W$.

## The Attempt at a Solution

Let W be a subspace of a vector space V, let y be in V and define the set $y + W = \{x \in V | x = y +w,\text{for some } w \in W\}$.
proof(←)
if $y \in W$ then any vector in $x \in y + W$ will satisfy $x=y+w$ such that $y,w \in W$. Since W is a subspace and is closed under addition, all the vectors in $y+W$ must also be in W, i.e. $y + W = W$.

proof(→)
...I'm stuck here. I'm thinking to do proof by contradiction? $y + W$ is a subspace and $y \not\in W$. I'm thinking to find a property it violates? But I don't know how to do this. Any hints on how I should proceed?

Similar to what Fresh wrote: you are translating a subspace by y . A subspace goes through the origin; if you shift it /translate it by y, you are adding y to each vector.

Yes. Maybe a bit complicated, but yes. The last part is relevant: If ##0\in y+W## then ##-y\in W## and all multiples as well, so ##(-1)\cdot (-y)=y \in W##, which is your contradiction. You don't need cases here.

But you can also skip the contradiction part, as you practically have concluded:
##y+W## subspace ##\Longrightarrow 0\in y+W \Longrightarrow -y \in W \Longrightarrow y\in W##
which had to be shown. The contradiction construction is a bit artificial.

I don't know if this is overkill, but the condition of going through the origin is "unstable", meaning it is delicate and broken with minor changes, such as, here, translation.But this is motivation and not a proof, obviously.

WWGD said:
the condition of going through the origin is "unstable",
Did I hear Zariski here?

The interesting part - which wasn't asked for - would have been to prove that ##\{\,y+W\,|\,y\in V\,\}## itself build a vector space.

fresh_42 said:
Did I hear Zariski here?

The interesting part - which wasn't asked for - would have been to prove that ##\{\,y+W\,|\,y\in V\,\}## itself build a vector space.

A spanning set, but definitely not eine minimalkeit

WWGD said:
Similar to what Fresh wrote: you are translating a subspace by y . A subspace goes through the origin; if you shift it /translate it by y, you are adding y to each vector.
Huh, didn't see it like that. Makes sense

fresh_42 said:
Yes. Maybe a bit complicated, but yes. The last part is relevant: If ##0\in y+W## then ##-y\in W## and all multiples as well, so ##(-1)\cdot (-y)=y \in W##, which is your contradiction. You don't need cases here.

But you can also skip the contradiction part, as you practically have concluded:
##y+W## subspace ##\Longrightarrow 0\in y+W \Longrightarrow -y \in W \Longrightarrow y\in W##
which had to be shown. The contradiction construction is a bit artificial.
It seems clear now thanks!

WWGD
WWGD said:
A spanning set, but definitely not eine minimalkeit
So? I was surprised that it didn't come. The author introduces cosets and then asked why a coset (other than ##W##) isn't a subspace? A bit lame, don't you think so?

fresh_42 said:
So? I was surprised that it didn't come. The author introduces cosets and then asked why a coset (other than ##W##) isn't a subspace? A bit lame, don't you think so?
Ok, good point, did not think of that.

## 1. What is the purpose of formulating a proof of subspaces?

The purpose of formulating a proof of subspaces is to provide a rigorous mathematical argument that a given set of vectors satisfies the definition of a subspace. This helps to establish the validity of claims made about the properties and characteristics of subspaces.

## 2. What are the key steps involved in formulating a proof of subspaces?

The key steps in formulating a proof of subspaces include:

1. Defining the subspace in terms of its properties and characteristics
2. Identifying the given set of vectors and showing that they satisfy the definition of a subspace
3. Proving that the set is closed under addition and scalar multiplication
4. Providing a counterexample to show that the set is not a subspace if necessary

## 3. What are some common techniques used in formulating a proof of subspaces?

Some common techniques used in formulating a proof of subspaces include:

• Direct proof: This involves using logical reasoning and mathematical operations to show that the set satisfies the definition of a subspace.
• Proof by contradiction: This involves assuming that the set is not a subspace and showing that it leads to a contradiction.
• Proof by induction: This involves proving that the set satisfies the definition of a subspace for a base case and then using mathematical induction to show that it holds for all other cases.

## 4. What are some challenges that may arise when formulating a proof of subspaces?

Some challenges that may arise when formulating a proof of subspaces include:

• Difficulty in identifying the correct properties and characteristics of the subspace
• Complexity in showing closure under addition and scalar multiplication
• Lack of understanding of the underlying concepts and definitions
• Difficulty in identifying and correcting errors in the proof

## 5. How can one improve their skills in formulating a proof of subspaces?

One can improve their skills in formulating a proof of subspaces by:

• Practicing regularly with various examples and exercises
• Studying and understanding the definitions and properties of subspaces
• Seeking help from a mentor or tutor when facing difficulties
• Collaborating with peers to discuss and solve problems together

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