# Why $V_a=\{a+x\mid x\in V\}$ is not a vector space?

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.

• It is just $V$. Any vector $v\in V$ can be written as $a+(v-a)$.
– lulu
Apr 10, 2020 at 12:18
• 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 $\lambda x$.
– lulu
Apr 10, 2020 at 12:20
• 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. Apr 10, 2020 at 12:22
• @lulu: So $V_a$ as defined is a vector space, right ? Apr 10, 2020 at 12:24
• @KaviRamaMurthy: Ok, but is it a vector space ? (because I don't see why the argument I've made doesn't work here). Apr 10, 2020 at 12:25

## 1 Answer

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