I am not sure, if my following proof is correct:
Let $V$ be an inner product space. Prove that if $V$ is the sum of pairwise orthogonal subspaces $U_1, \ldots U_n$, the sum must be direct; $V = \bigoplus _{i = 1}^{n} U_i$.
My proof:
Since $V = $span$(U_1, \ldots U_n)$, we only have to show that the subspaces are pairwise disjoint.
Let $v \in V$ be a vector with $v \not= 0$ and $v \in U_i \cap U_j$ for some $i, j \in \{1, \ldots, n\}$ with $i \not= j$. Then $v \in U_i$ and $v \in U_j$ follows. Since we require $U_i \perp U_j$, every vector from $U_i$ has to be orthogonal to every vector in $U_j$, so especially we would have $v \perp v$, which is only possible if $v = 0$, leading us to a contradiction. $\square$
My Question:
Is my method correct or do I have to prove that $U_1$ and span$(U_1, U_2)$ and span$(U_1, U_2, U_3)$ and so on are disjoint?
Best Answer
Pairwise disjoint of $(U_1,\ldots,U_n)$ doesn't imply direct sum of the sum $\sum U_i$ as you can see a counterexample of three different lines in $\Bbb R^2$. A simple way to prove the result is to take $x_i\in U_i$ such that $x_1+\dots+x_n=0_V$ and prove that $x_i=0_V$ for all $i$. Indeed, for all $i$ $$0=\langle x_i,x_1+\cdots+x_n\rangle=\langle x_i,x_i\rangle\implies x_i=0_V$$