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# System prepended metadata

title: Untitled

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want to show: $\langle\{m+U\mid m\in M\}\rangle = \{\{v+U\} |v\in V\}$


"$\subseteq$"

let $A\in\langle\{m+U\mid m\in M\}\rangle$. Then there are $a_m\in\mathbb{R}\setminus 0$ and $\tilde{M}\subseteq M$ such that $A=\sum\limits_{m\in \tilde{M}}a_m*\{m+U\} = \sum\limits_{m\in \tilde{M}}\{a_m*(m+U)\}$

note that $a_m*U=U$ for $a_m\neq 0$ and $U$ a vector space. Thus:

$A= \sum\limits_{m\in \tilde{M}}\{a_m*m+U\}= \sum\limits_{m\in \tilde{M}}(\{a_m*m\}+U)$

As $U+U=U$ because $U$ is a vector space:

$A=(\sum\limits_{m\in \tilde{M}}\{a_m*m\})+U$

let $v':=\sum\limits_{m\in \tilde{M}}a_m*m\in V$, then

$A=\{v'\}+U=v'+U\in \{\{v+U\} |v\in V\}$


"$\supseteq$"

let $A=v+U$ for some $v\in V$. Because $V=\langle M\rangle$ we can write $v=\sum\limits_{m\in\tilde{M}}a_m*m$ with $\tilde{M}\subseteq M$ and $a_m\neq0$.Then:

$A=(\sum\limits_{m\in\tilde{M}}a_m*m)+U=\sum\limits_{m\in\tilde{M}}(a_m*m+U)=\sum\limits_{m\in\tilde{M}}(a_m*m+a_m*U)$ 

$=\sum\limits_{m\in\tilde{M}}a_m*\{m+U\}\in\langle\{m+U\mid m\in M\}\rangle$