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mathmari
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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:
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:
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