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Let $V$ a $K-$vector space of finite dimension and $a\in V\setminus \{0\}$. Why $$V_a=\{x+a\mid x\in V\}$$ is not a vector space ? In my correction it's written that $0\notin V_a$, it's not stable for $+$ neither for multiplication by a scalar.

We have that $0=a-a\in V_a$, and thus $0\in V_a$. Moreover, $$(a+x)+(a+y)=a+(a+x+y)\in V_a$$ and $$\lambda(a+x)=\lambda a+\lambda x=(1+h)a+\lambda x=a+(ha+\lambda x),\in V_a$$ where $h\in\mathbb R$.

So, I don't understand there arguments.

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  • $\begingroup$ It is just $V$. Any vector $v\in V$ can be written as $a+(v-a)$. $\endgroup$
    – lulu
    Apr 10, 2020 at 12:18
  • $\begingroup$ I expect that you (or your reference) mean to refer to a set of the form $\{\lambda x+a\}$ where $x$ is a fixed vector, $\lambda$ is a variable scalar and $a$ is a fixed vector not of the form $\lambda x$. $\endgroup$
    – lulu
    Apr 10, 2020 at 12:20
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    $\begingroup$ If $M$ is a subspace of $V$ and $a \notin M$ then $\{x+a:x\in M\}$ is not subspace. I suspect that you are mis-quoting this result. $\endgroup$ Apr 10, 2020 at 12:22
  • $\begingroup$ @lulu: So $V_a$ as defined is a vector space, right ? $\endgroup$
    – Walace
    Apr 10, 2020 at 12:24
  • $\begingroup$ @KaviRamaMurthy: Ok, but is it a vector space ? (because I don't see why the argument I've made doesn't work here). $\endgroup$
    – Walace
    Apr 10, 2020 at 12:25

1 Answer 1

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Your arguments are correct. $V_a$ is vector space. you have proved that it is not empty and closed under vector addition and scalar multiplication

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