# Formulation of a proof of subspaces

• 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?

## 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?

## 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.

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.

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

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

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
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?

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.