Few problems about vector being isomorphic 
Let $U$,$V$,$W$ be vector spaces
$1$
.How do I prove   that  $U \times ( V \times W)$ is isomorphic to $(U \times V) \times W$
Let $V$ be vector space
$2$.How do I prove  that  $V\times {0}$ is isomorphic to $V$

I am not getting the concept for isomorphic..
what conditions do I have to check for vector being isomorphic to another vector?
 A: Try this :
Define $T:(U\times V)\times W\to U\times (V\times W)$ by $T((u,v),w)=(u,(v,w))$
Check its linear and bijective .
For the second one $T(v)=(v,0)$
A: For a vector space $V$ to be isomorphic to another vector space $W$ (where they are both over the same field $F$) you need to show the existence of a bijective linear transformation between them. 
Now an arbitrary element of $U\times (V\times W)$ is $(u,(v,w))$. A bijective linear transformation from $U\times (V\times W)$ to $(U\times V)\times W$ is given by $(u,(v,w))\to ((u,v),w)$. Similarly a bijective linear transformation from $V\times\{0\}$ to $V$ is given by $(v,0)\to v$. You should show all the properties of a bijective linear transformation for both of these, namely that these maps are $1-1$ onto functions and are also a linear transformations.
A: Two vector spaces over same field are said to be isomorphic if there is a bijective linear transformation between them. That means there should be one one and onto linear map between them.
Now for your first case, we can define $T_1:U\times (V\times W)\to (U\times V)\times W$ by $T_1(u,(v,w))=((u,v),w)$
You can see that this is a linear map. 
Also this is one one since $((u,v),w)=((u',v'),w') \Rightarrow u=u',v=v',w=w'$ 
It is onto because for $((u,v),w) \in (U\times V)\times W$ we have $(u,(v,w)) \in U\times (V\times W)$ with  $T_1(u,(v,w))=((u,v),w)$
Hence this map is a bijective linear transformation.
For the second case you just try with $T_2:V\times{0} \to V$ by $T_2((v,0)=v$
It will work
