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i am having problems with the following problem:

Let $V$ be a $K$-vector space and $U \subseteq V$ a linear subspace. Determine all vectors $v \in V$ so that the set $v + U$ is a subspace of $V$.
This is my idea of solving it:
If $v + U$ is a subset of $V$ then it is $(v + u_{1}) + (v + u_{2}) \in v + U$ for all vectors $u_{1},u_{2} \in U$, wich means there exists $u_{3} \in U$ so that $(v + u_{1}) + (v + u_{2}) = (v + u_{3})$ $\Rightarrow 2v + u_{1} + u_{2} = v + u_{3}$
$\Rightarrow v = u_{3} - u_{1} - u_{2} $
Since $U$ is a subspace and $u_{3} - u_{1} - u_{2}$ is still in $U ~~~~\Rightarrow v \in U$

But it seems fairly obvious to me that i can add any vector $v \in U$ to $U$ so that it remains a subspace, aren't there any other vectors $v$ that I am missing? If not is my proof correct then?

Thx in advance.

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    You have it right. To be a subspace we need $\vec 0\in \vec v+U$ which implies $-\vec v \in U$ so $\vec v \in U$. – 2017-02-23
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    "But it seems fairly obvious to me that I can add any vector $v$ to $U$ so that it remains a subspace"; sure, it seemed pretty obvious to me too, but in fact its a mistake; if you think of $U$ as a plane through the origin, then adding a vector $v$ to it shifts the plane in the direction of $v$; but whilst it remains a plane - which at least for me is the source of the confusion - it's no longer going through the origin; this means it can't include the zero vector and so it can't be a vector space. – 2017-02-23
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    when the vector $v$ is in the plane, then you;re simply translating the plane *within* the plane, so it stills remains on the origin; this matches your result. – 2017-02-23
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    You are quoting me wrong, i said you can add and vector $v \in U$ to $U$ so that it remains a subspace, it is clear that this does not work for just any vector. – 2017-02-23

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