# $\dim(U)=\dim(U_1)+\dots+\dim(U_n)\implies$ every $x\in U$ has a unique representation $x=u_1+\dots+u_n$, with $u_i\in U_i$

Let $U$ be a finite dimensional vector space, and $U=U_1+U_2+\dots+U_n$ with $U_i$ subspaces, show that if $\dim(U)=\dim(U_1)+\dots+\dim(U_n)$ then every $x\in U$ has a unique representation $x=u_1+\cdots+u_n$, with $u_i\in U_i$.

for $n=2$ assume the converse, $x=u_1+u_2=u'_1+u'_2$, then $0=(u_1-u'_1)+(u_2-u'_2)\in U_1\cap U_2\Rightarrow(u_2-u'_2)\in U_1\cap U_2$, so by dimension formula $\dim(U_1+U_2)=\dim(U_1)+\dim(U_2)-\dim(U_1\cap U_2)$, and $U_1\cap U_2$ is nonempty, but what if $n\ge2$ then the dimension formula looks complicated by inclusion exclusion etc.

For all $i$, let $B_i$ be a basis of $U_i$. Since $U = \sum_i U_i$, you have that $\cup_i B_i =B$ generates $U$. Moreover, by your assumption, you have $$|\cup_i B_i| \le \sum_i |B_i| = \sum_i \dim U_i = \dim U \le |\cup_i B_i|$$ hence equality holds. This means that $B$ is a basis of $U$, and that all the $B_i$s are pairwise disjoint. This concludes the proof.
• Yes, I really mean the union. I mean an union of sets, not of spaces. For example, if $U_1$ has basis $B_1 = \{ v,w \}$, and $U_2$ has basis $B_2 = \{ x,y,z \}$, then $U_1 + U_2$ is generated by $B_1 \cup B_2 = \{ v,w,x,y,z\}$ (but in general it is not a basis!). Oct 29, 2015 at 12:49
Using induction: assume that $$\dim U = \dim U_1 + \cdots + \dim U_n \implies { \rm every }\,x\ldots$$ then, assuming $$\dim U = \dim U_1 + \cdots + \dim U_{n+1}$$ and defining $V =U_1 + \cdots + U_n$: $U = V + U_{n+1}$. $$\dim U \le \dim V + \dim U_{n+1} \le \dim U_1 + \cdots + \dim U_{n} + \dim U_{n+1} = \dim U$$ so every inequality is an equality and in particular: $$\dim V = \dim U_1 + \cdots + \dim U_{n}$$ (so every $v\in V$ writes itself uniquely as a sum $u_1 + \cdots + u_n$ where $u_i\in U_i$) and $$\dim U = \dim V + \dim U_{n+1}$$so every $u\in U$ writes itself uniquely as $u = v + u_{n+1}$, $v\in V$ and $u_{n+1}\in U_{n+1}$.