Prove that Euclidean space is the direct sum of $U$ and $U^\perp$, where $U$ and $U^\perp$ are the subspaces of $E$ Because we know that $\dim E=\dim U+\dim U^\perp-\dim(U\cap U^\perp)$ is it enough to prove that the intersection of $U$ and $U^\perp$ is the null-space? That would mean that $\dim E=\dim U+\dim U^\perp$ and so would $E$ be equal to $U+U^\perp$.
In my lictures they proved like this:
The've first found the orthogonal base for $U$ and filled the base up so that they get a orthogonal base for $E$. Then they've written each vector $x\in E$ like $x=y+z$,where $y\in U$,$z\in U^\perp$ so that they get $E=U+U^\perp$ and then they've proved that $U \cap U^\perp = \{0\}$ which led them to $E=U\bigoplus U^\perp$.
I don't get why they've first had to prove that $E=U+U^\perp$ and then to prove that their intersection is the null-space?
My guess is that we are dealing here with a scalar product. Am I right? Them yes, your argument is correct: if you prove that $U\cap U^\bot=\{0\}$, then you shall have proved that $E$ is the direct sum of $U$ and $U^\bot$ assuming that you know that $E=U+U^\bot$.
You have $U\cap U^\bot=\{0\}$ because if $v\in U\cap U^\bot=\{0\}$, then $\langle v,v\rangle=0\Longrightarrow\|v\|=0\Longrightarrow v=0$.