Check My Proof - $V = \sum_{i = 1}^{n} U_i \implies V = \bigoplus_{i = 1}^{n} U_i$ if $U_i \bot U_j \ \forall i,j \in \{1, \ldots, n\}$

Question

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?

Answer 1

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 $i$ $$0=\langle x_i,x_1+\cdots+x_n\rangle=\langle x_i,x_i\rangle\implies x_i=0_V$$

Answer 2

As I pointed out in the comments, you need to prove a stronger condition. Here is a possible approach: let $j \in \{1, \ldots, n\}$, and lets show that $U_j \cap \bigoplus_{i \neq j}U_i = 0$. In effect, let $u \in U_j$ such that $u = \sum_{i \neq j}u_i$ with each $u_i \in U_i$. Now,

$$ \langle u,u \rangle = \langle u,\sum_{i \neq j}u_i\rangle = \sum_{i \neq j}\langle u,u_i\rangle = 0 $$

because $\langle u , u_i \rangle = 0$ for each $i$, since $U_j \perp U_i$ for $i \neq j$. Thus, $u = 0$.

Answer 3

A little modification of your proof simply using induction leads the following:

Induction start $n=1$: Clear since $\{0\} \cap U_1 = \{0\}$.

Induction step: Assume for any $n>1$ the Sum $V_{n-1} = \bigoplus^{n-1}_{k=1}U_k$ is direct. Let $v\in V_{n-1} \cap U_{n}$ with $v \neq 0$. But then, $v\in V_{n-1}$ and $v \in U_{n}$ and since $V_{n-1} \perp U_n$ we have $\left< v,v\right>=0$, so $v=0$. Contradiction. Henceforth, the sum $V_{n-1} \oplus U_n$ is direct and so is $V_n = \bigoplus^n_{k=1} U_k$. $\blacksquare$