# Prove by induction that the direct sum of subspaces of a vector space...

A question asked by my math professor:

Prove by induction that if $$W_1, W_2, ... , W_n \subseteq W$$ are subspaces of a vector space W over F, then $$W = W_1 \oplus W_2 \oplus ...\oplus W_n$$ if and only if

$$W = W_1 + W_2 + ... + W_n$$ and $$W_i \cap (W_1 + W_2 + ... + W_{i-1}+W_{i+1}+...+W_n) = \{0\}.$$

I demonstrated a base case with $$W_1, W_2.$$

However, what would my inductive hypothesis be? It confuses me. For example, if I say $$W = W_1 \oplus W_2 \oplus ...\oplus W_k,$$ I can't say $$W = W_1 \oplus ... \oplus W_{k+1},$$ unless $$W_{k+1} = \{0\}.$$

I think it should be:

$H(n)$: "For any subspace $F$ of $W$ and for any $k$ subspaces of $F$ with $1≤k≤n$, if $F = W_1 + ...+ W_{k}$ and the intersection hypothesis is also true then $F = W_1 \oplus ...\oplus W_k$".

For the induction step ($n \to n+1$), let $V = W_1 \oplus ...\oplus W_{n-1}$, apply the induction hypothesis to $V$ with $k = n-1$ and then write $W= V \oplus W_n$ and apply the induction hypothesis to $W$ with $k=2$.

• I need clarification. If I may, am I recursively going down the list. That is, I begin with n and will pair n-1 and n, not n and n+1? So, I say what about W_1 \oplus W_2 \oplus ... \oplus W_n-1? Feb 2 '18 at 6:20
• I have edited my post. The induction hypothesis is quite complicated but it seems necessary otherwise we are stuck in the induction step.
– user371663
Feb 2 '18 at 6:29

Inductive step:

Assume there exists some $k\in \mathbb{N}$ such that

"For any subspaces $W_1,W_2,\dots,W_k\subseteq W$, we have the equivalence:

$W=W_1\oplus W_2\oplus \dots \oplus W_k$ iff $W=W_1+ W_2+ \dots + W_k$ and $W_i\cap(W_1+W_2+\dots+W_{i-1}+W_{i+1}+\dots+W_k)=\{0\}$"

Then your step is to prove that:

"For any subspaces $W_1,W_2,\dots,W_{k+1}\subseteq W$, we have the equivalence:

$W=W_1\oplus W_2\oplus \dots \oplus W_{k+1}$ iff $W=W_1+ W_2+ \dots + W_{k+1}$ and $W_i\cap(W_1+W_2+\dots+W_{i-1}+W_{i+1}+\dots+W_{k+1})=\{0\}$"