$\dim V \geq \dim U$, then there exists a linear injection from $U$ to $V$. 
Let $U$ be a $K$ vector space with $\dim U=n$. I have to show that for
  all $r\in \mathbb{N}$ with $n\leq r$, there is a $K$ vector space $V$
  with $\dim V=r$ and a linear injection $U\hookrightarrow V$.

I have no idea how to go about this, can someone help?
 A: Hint: Say we already have a $K$-vector space, $V$, of dimension $r$ somehow. Suppose $u_1,u_2,\dots,u_n$ form a basis for $U$ over $K$ and $v_1,v_2,\dots,v_r$ form a basis for $V$ over $K$. To get a well-defined linear map, $T$ over $U$, it is enough to define $T$ at each of the values $u_1,u_2,\dots,u_n$. More specifically, to get a well-defined linear injection, $T$ over $U$, it is enough to define $T$ at each of the values $u_1,u_2,\dots,u_n$ so that $T(u_i)\neq T(u_j), \;\forall\; i\neq j$. Using the facts given, can you see how to get an injection $:U\rightarrow V$?  
Full Answer: 

 As in Osama Ghani's answer, set $V=K^r$. We simply define $i:U\rightarrow V$ by $i(u_k)=v_k\;\forall\; 1\leq k\leq n$. Since $r\geq n$, this is well-defined. Also, if $k\neq j$ then $v_k\neq v_j$ (as the list of $v_k$s are linearly independent), so $i(u_k)\neq i(u_j)\;\forall\; k\neq j$. We can extend $i$ to all of $U$ by supposing that it satisfies $i(a+b)=i(a)+i(b)\;\forall\; a,b\in U\;$, $\;i(\gamma\cdot a)=\gamma\cdot i(a)\;\forall\; a\in U,\;\forall\; \gamma\in K$, and using the fact that each element in $U$ can be expressed as a unique linear combination of the elements $u_1,u_2,\dots,u_n$. This map is clearly injective since if for any $u=a_1u_1+a_2u_2+\dots+a_nu_n\in U$ we have $i(u)=\overrightarrow 0$, then $a_1i(u_1)+a_2i(u_2)+\dots+a_ni(u_n)=\overrightarrow 0$ which means $a_1v_1+a_2v_2+\dots+a_nv_n=\overrightarrow 0$. As the list of all $v_k$s is linearly independent so is the list of the first $n$ $v_k$s. So we have that all $a_i$s are $0$, i.e. $u=(0)u_1+(0)u_2+\dots+(0)u_n=\overrightarrow 0$, making $i$ injective.   

(Osama Ghani's comment referenced)
