Let $(V, K)$ be a vector space and $W ⊂ V$ a subspace. A subset $S ⊂ V$ is called a $W$-affine subspace of $V$ if the following holds $∀s, s ∈ S, s − s ∈ W$ and $∀s ∈ S, ∀w ∈ W, s + w ∈ S$.

  1. Let S and T be W-affine subspace of V and c ∈ K. Define

$$S + T = \lbrace s + t \space | \space s ∈ S, t ∈ T\rbrace$$

$$cT = \begin{cases} \lbrace ct \space | \space t ∈ T\rbrace \space \text{if} \space c= 0\\ W \space \text{if} \space c = 0. \end{cases}$$ Show that $S+T$ and $cT$ are again $W$-affine subspaces of $V$.

  1. Show that the above operations turn the set of all $W$-affine subspaces of $V$ into a vector space that we will denote by the symbol $V/W$ and we call it the quotient space. Note that vectors in the quotient space are $W$-affine subsets of $V$.
  2. Show that if $v ∈ V$ then $v + W = \lbrace v + w \space|\space w ∈ W\rbrace$ is a $W$-affine subspace of $V$. Furthermore, show that for any $W$-affine subspace $S ∈ V/W$ there exists a $v ∈ V$ such that $S = v + W$.

Hello there. I know this question is quite long, but the parts are all related, so just bear with me please. I attempted parts $1$ and $2$, but I was not sure about these attempts and I completely do not know how to approach number $3$. Any help will be greatly appreciated.

My attempts:

  1. Since $S$ and $T$ are $W$-affine subspaces (by definition)

$s-s' \in W \space \forall \space s, s' \in S, t-t' \in W \space \forall \space t, t' \in T$


$s-s' \in W \implies s-s'=w_1\in W$ and $t-t' \in W \implies t-t'=w_2 \in W$

By definition of $S + T$;

$$S + T = \underbrace{(s - s')}_{\in \space S} + \underbrace{(t - t')}_{\in \space T}\\ = (s - s') + (t - t') = w_1 + w_2\\ =(s + t) - (s' + t') = w_1 + w_2$$

Since $W$ is a subspace, $w_1 + w_2 \in W$,

$$\implies S + T = (s + t) - (s' + t') \in W$$

Also, since $S$ and $T$ are $W$-affine subspaces, $s + w_1 \in S$ and $t + w_2 \in T$ $$\therefore S + T = \underbrace{(s + w_1)}_{\in \space S} + \underbrace{(t + w_2)}_{\in \space T}\\ =(s + t) + \underbrace{(w_1 + w_2)}_{\text{any}\space w \space \in \space W}$$ $$\implies S + T = (s + t) + w$$

And therefore, $S + T$ is again a $W$-affine subspace of $V$.

Similar process for $cT$ ending up with: $$cT =c(t - t') \in W, \text{and}\space cT = ct + w$$

  1. I defined addition and scalar multiplication as follows $$+ : S + T = (s + t) + (t + w) = (s + t) + w$$ $$\cdot : cT = c(t + w) = (ct) + w$$ and then just showed (using those definitions) that they satisfy the properties of a vector space: commutative, associative, additive identity, additive inverse, multiplicative inverse, distributivity of $\cdot$ over $+$ and addition distribution property and thus $V/W$ forms a vector space.

  2. Was thinking it'd somehow look like my approach for number $1$?? Not entirely sure how to approach it.

Thanks for taking the time to read/help me :)


1 Answer 1


1.- The proof goes as follows (see that for this part you don't need them to be subspaces):

Let $s_1+t_1$ and $s_2+t_2$ arbitrary elements in $S+T$. Then $(s_1+t_1)-(s_2+t_2) = (s_1-s_2)+(t_1-t_2)$. Since both sets are $W$-affine there exist $w_s$ and $w_t \in W$ such that $s_1-s_2=w_s$ and $t_1-t_2=w_t$. Since $W$ is a subspace, this implies that $(s_1+t_1)-(s_2+t_2)\in W$, hence $S+T$ is a $W$-affine.

2.- With 1 you already defined both operations, after this you should verify the distributivity, find the inverses and the identity sets. Hint: $W$ itself is $W$-affine.

Edit: Observe that $S+W=S$ for every $W$-affine set $S$. So this is your zero vector in $V/W$. For the inverse, you can show that $S+S=W$ (since $S$ is a subspace this is equivalent to taking $(s+W)+((-s)+W)=W$).

3.- I'll leave the first part to you. For the second part one contention follows immediately from the definition. For the other see that $S\subseteq s+W$ for any $s\in S$. Show it doesn't matter which $s$ you choose ($s+W=s'+W$).

  • $\begingroup$ Firstly thank you. Okay I understand the corrections to $1$ - the way you suggested is indeed more logical than what I did, doing $cT$ now. For $2$ does that mean I should use the definitions given in part $1$ when verifying distributivity, finding the inverses and the identities? Cause I verified the distributivity and found the inverse and identity with the definitions I put forward. $\endgroup$
    – Zhoe
    Sep 29, 2014 at 1:51
  • $\begingroup$ a) You should also prove the other condition ($s+w\in S$), in my post I only proved the first one. b) Well, I think that's why the first part of the exercise is to define a sum and a scalar multiplication ;) (also I think what you did isn't right, I don't think you're proving what you intend to). $\endgroup$
    – hjhjhj57
    Sep 29, 2014 at 2:00
  • $\begingroup$ Okay, so would the way how I proved the second part of $1$ be correct, or is that also incorrect?? I was pretty convinced that what I did was incorrect, I was sort of cheating, cause I was entirely unsure of how to start.. $\endgroup$
    – Zhoe
    Sep 29, 2014 at 2:07
  • 1
    $\begingroup$ Thank you! Yes that is much better. & I did realize that, but I couldnt conceive a nicer way of writing it. Okay so for $cT$ this is what I came up with. When $c=0$, $cT$ is $W \in W$ so that case is ok. If $c \ne 0$ then for any arbitrary $t, t' \in T$, $ct - ct' = c(t-t') \in W$ since $T$ is $W$-affine, then $\exists w_t$ such that $t-t'=w_t$ and since $W$ is a subspace $\implies ct - ct' \in W$. $\endgroup$
    – Zhoe
    Sep 29, 2014 at 2:36
  • 1
    $\begingroup$ For the second one, if $c=0$, $cT$ is $W$. Take an arbitrary $w \in W$, $\implies t=0$ and therefore $t+w=0+w \in T$. If $c\ne 0$, let $t \in T$, since $T$ is $W$-affine, for any arbitrary $w$, $t+w \in T$, $\implies c(t+w) = ct' \in T$. $\endgroup$
    – Zhoe
    Sep 29, 2014 at 2:52

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