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Statements about subspaces

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
Hey!! :giggle:

Let $V$ be a $\mathbb{R}$-vector space, let $x,y\in V$ and let $U,W\leq_{\mathbb{R}}V$ be subspaces of $V$.
Show that :
(a) If $(x+U)\cap (y+W)\neq \emptyset$ and $z\in (x+U)\cap (y+W)$ then $(x+U)\cap (y+W)=z+(U\cap W)$.
(b) The following statements are equivalent:
(i) $U=W$ and $x-y\in U$.
(ii) $x+U=y+W$.


I have done the following :

(a) let $z\in (x+U)\cap (y+W)$. That means that $z\in x+U$ and $z\in y+W$. So we have that $z=x+u$, for $u\in U$ and $z=y+w$, for $w\in W$.
How can we continue?

(b) $(i)\Rightarrow (ii)$ :
Since $x-y\in U$ we have that $x-y=u\Rightarrow x=y+u$. Then we have that $x+U=y+u+U=y+U$. Since $U=W$ we have that $y+U=y+W$. So we get $x+U=y+W$.

$(ii) \Rightarrow(i)$ :
Since $x+U=y+W$ we have that $x+u=y+w$, for $u\in U, w\in W$. Then we get $x-y=w-u$. How can we continue?


:unsure:
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
Hey mathmari !!

Suppose $z'$ is in the left hand set as well. Can we then prove it must be in the right hand set?
That is, can we prove that $z-z'$ is in $U\cap W$? 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
Suppose $z'$ is in the left hand set as well. Can we then prove it must be in the right hand set?
That is, can we prove that $z-z'$ is in $U\cap W$? 🤔
Suppose $z'\in (x+U)\cap (y+W)$.
So we have that $z=x+u_1$, $z'=x+u_2$, for $u_1, u_2\in U$ and $z=y+w_1$, $z'=y+w_2$, for $w_1, w_2\in W$.
Then $z'-z=u_2-u_1\in U$ and $z'-z=w_2-w_1\in W$, so $z'-z\in U\cap W$.
This means that $z'\in z+U\cap W$, i.e. $(x+U)\cap (y+W)\subseteq z+U\cap W$.

For the other direction :
Let $z'\in z+U\cap W$, with $z\in (x+U)\cap (y+W)$. Do we take now a specific $z$ ? I mean that it is of the form $x+u$ and $y+w$ ?

:unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
Suppose $z'\in (x+U)\cap (y+W)$.
So we have that $z=x+u_1$, $z'=x+u_2$, for $u_1, u_2\in U$ and $z=y+w_1$, $z'=y+w_2$, for $w_1, w_2\in W$.
Then $z'-z=u_2-u_1\in U$ and $z'-z=w_2-w_1\in W$, so $z'-z\in U\cap W$.
This means that $z'\in z+U\cap W$, i.e. $(x+U)\cap (y+W)\subseteq z+U\cap W$.
Good. (Nod)

For the other direction :
Let $z'\in z+U\cap W$, with $z\in (x+U)\cap (y+W)$. Do we take now a specific $z$ ? I mean that it is of the form $x+u$ and $y+w$ ?
We can take the same $z$ as before, since it is given in the problem statement.
$z'$ must then be of the form $z'=z+u'=z+w'$ with $u'=w'$ in both $U$ and $W$. 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
We can take the same $z$ as before, since it is given in the problem statement.
$z'$ must then be of the form $z'=z+u'=z+w'$ with $u'=w'$ in both $U$ and $W$. 🤔
Since $z\in (x+U)\cap (y+W)$ we have that $z=x+u$ and $z=y+w$. So we get that $z'=x+u+u'\in x+U$ (since $u,u'\in U$ and so $u+u'\in U$ since $U$ is a subspace) and $z'=y+w+w'\in y+W$ (since $w,w'\in W$ and so $w+w'\in W$ since $W$ is a subspace).
That means that $z'\in x+U \ \cap \ y+W$.

Is that correct? :unsure:
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
As for question (b) :

$(i)\Rightarrow (ii)$ :
Since $x-y\in U$ we have that $x-y=u\Rightarrow x=y+u$. Then we have that $x+U=y+u+U=y+U$. Since $U=W$ we have that $y+U=y+W$. So we get $x+U=y+W$.

$(ii) \Rightarrow(i)$ :
Since $x+U=y+W$ we have that $x+u\in y+W$, for every $u\in U$ and $y+w\in x+U$, for every $w\in W$. Since $U$ and $W$ are subspaces $0$ is contained in both sets. So we get $x+0\in y+W \Rightarrow x-y\in W$ and $y+0\in x+U \Rightarrow y-x\in U \Rightarrow x-y\in U$.
So $x-y\in U\cap W$.

Is that correct so far? How do we continue to show that these sets are equal? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
Since $z\in (x+U)\cap (y+W)$ we have that $z=x+u$ and $z=y+w$. So we get that $z'=x+u+u'\in x+U$ (since $u,u'\in U$ and so $u+u'\in U$ since $U$ is a subspace) and $z'=y+w+w'\in y+W$ (since $w,w'\in W$ and so $w+w'\in W$ since $W$ is a subspace).
That means that $z'\in x+U \ \cap \ y+W$.

Is that correct?
Yep. (Nod)

As for question (b) :

$(i)\Rightarrow (ii)$ :
Since $x-y\in U$ we have that $x-y=u\Rightarrow x=y+u$. Then we have that $x+U=y+u+U=y+U$. Since $U=W$ we have that $y+U=y+W$. So we get $x+U=y+W$.

$(ii) \Rightarrow(i)$ :
Since $x+U=y+W$ we have that $x+u\in y+W$, for every $u\in U$ and $y+w\in x+U$, for every $w\in W$. Since $U$ and $W$ are subspaces $0$ is contained in both sets. So we get $x+0\in y+W \Rightarrow x-y\in W$ and $y+0\in x+U \Rightarrow y-x\in U \Rightarrow x-y\in U$.
So $x-y\in U\cap W$.

Is that correct so far? How do we continue to show that these sets are equal?
Looks correct to me. (Nod)

When in doubt, the usual way to show sets are equal, is by contradiction.
Suppose the sets are not equal, then there must be... 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
When in doubt, the usual way to show sets are equal, is by contradiction.
Suppose the sets are not equal, then there must be... 🤔
We assume that $U\neq W$, then we have for example $U\subset W$. So there is a $w'\in W$ such that $w'\notin U$. But for this element it holds that $y+w'\in x+U$, so $-(x-y)+w'\in U$.
It holds that $x-y\in U$, so also $-(x-y)\in U$.
It also holds that $w'\in U$, a contradiction.

So it must be that $U=W$ and that $x-y\in U\cap W=U$.


Is that correct? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
We assume that $U\neq W$, then we have for example $U\subset W$. So there is a $w'\in W$ such that $w'\notin U$. But for this element it holds that $y+w'\in x+U$, so $-(x-y)+w'\in U$.
It holds that $x-y\in U$, so also $-(x-y)\in U$.
It also holds that $w'\in U$, a contradiction.

So it must be that $U=W$ and that $x-y\in U\cap W=U$.

Is that correct?
I don't think we should state that "we have for example $U\subset W$. So there is a $w'\in W$ such that $w'\notin U$."
There are multiple relationships that $U$ and $W$ can have.
Instead we can state that if $U\ne W$ that there must be either an $u'$ in $U$ that is not in $W$ and/or there must be a $w'$ in $W$ that is not in $U$.
You continued with the second possibility, which is fine.
We should still state that the first possibility also leads to a contraction, for instance due to symmetry. 🤔
 

mathmari

Well-known member
MHB Site Helper
Apr 14, 2013
4,713
I don't think we should state that "we have for example $U\subset W$. So there is a $w'\in W$ such that $w'\notin U$."
There are multiple relationships that $U$ and $W$ can have.
Instead we can state that if $U\ne W$ that there must be either an $u'$ in $U$ that is not in $W$ and/or there must be a $w'$ in $W$ that is not in $U$.
You continued with the second possibility, which is fine.
We should still state that the first possibility also leads to a contraction, for instance due to symmetry. 🤔
Ahh I see! Thank you!! (Smile)