How to show that a sum of subspaces can be made into a direct sum using subspaces which are subsets of each original subspace

Question

Let $W_1,\ldots,W_n$ be subspaces of a vector space $V$. Prove that there exist subspaces $U_1 \subseteq W_1, \ldots , U_n \subseteq W_n$ such that $W_1 + \cdots + W_n = U_1 \oplus \cdots \oplus U_n$. (direct sum)

Should I start by defining a basis for each $W_i$? I could see where I could obtain a basis for $W_1 + \cdots + W_n$ by taking the union of the bases for each $W_i$ and then eliminating dependent elements.

I know that for $W_1 + \cdots + W_n$ to be the direct sum of the $U_i$'s, I need the intersection of each $U_i$ with the rest to be zero.

Just not sure how to rigorously start this proof or how to go about defining the $U_i$'s.

Answer 1

I will first prove the theorem for $n=2$. Intersection $$V=W_1 \cap W_2$$ of vector spaces $W_1$ and $W_2$ is also a vector space. If $V=\{0\}$, the sum $W_1 \oplus W_2$ is already direct and there is nothing more to prove. Othewise, let $$\{v_1,\ldots,v_k\}$$ be a basis of vector space $V$. We can always augment this basis with vectors $v_{k+1},\ldots,v_m$ to a basis $\{v_1,\dots,v_m\}$ of $W_2$, and then $$W_1+W_2=W_1 \oplus \operatorname{span}(v_{k+1},\ldots,v_m)$$ where $\operatorname{span}(v_{k+1},\ldots,v_m) \subseteq W_2$.

Using construction above, you can easily prove the theorem for $n>2$ by a mathematical induction.