# Need an example or contradiction.

If $A$ is a vector space then is it always true that $A+A=2A$. I know that it's not true but whats the use of saying no when i can't prove it.

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Over a field of characteristic $2$ it can fail. Otherwise not, since each is equal to $A$. – André Nicolas Jun 5 '12 at 15:17
@AndréNicolas : i.e , either $A+A= A or 2A$ is it ? – Theorem Jun 5 '12 at 15:20
$A=A+\{0_A\}\subseteq A+A\subseteq A$, where the second inclusion is due to $A$ being an abelian group, and so, for every $a,b\in A$, $a+b\in A$. – Josué Tonelli-Cueto Jun 5 '12 at 15:27
what do mean by the addition of tow vector spaces "$A+A$" ? – Abdelmajid Khadari Jun 5 '12 at 15:32
@AbdelmajidKhadari I take it to mean the set $\{x+y|x \in A,y \in A\}$. – Ravi Jun 5 '12 at 15:42

We always have $A+A=A$, since a vector space is closed under addition. To fill in all details, note first that $A\subseteq A+A$, since any vector $v$ can be expressed as $v$ plus the zero-vector. And $A+A\subseteq A$, since if $u\in A$ and $v\in A$ then $u+v\in A$.
Now look at $2A$, which is the set of all vectors of the form $(1+1)v$, where $v\in A$. It is clear that $2A\subseteq A$, since if $v\in A$ then $2v\in A$.
If $1+1$ is invertible in the underlying field of the vector space, then we also have $A\subseteq 2A$. For if $v\in A$, and $k$ is the multiplicative inverse of $1+1$ in the field, then $v=(1+1)(kv)$.
If $1+1$ is not invertible in the underlying field, we can have $(1+1)A\ne A$. For example, let $A$ be the $1$-dimensional (or $n$-dimensional where $n\ge 1$) space over $\mathbb{Z_2}$. Then $(1+1)A$ is just the space consisting of the zero-vector, so $(1+1)A$ is not equal to $A$.