If $E = F \bigoplus G$ and $F$ and $G$ are orthogonal, then why is there $x_F+x_G=x \in F^{\perp}$?

Question

In a proof I am trying to understand, this fact is stated:

Let $F$ and $G$ be subspaces of $E$ such that:
- $F$ and $G$ are orthogonal
- $E=F \bigoplus G$
Let $x \in F^{\perp}$, then $\exists x_F \in F, \exists x_G \in G$ such that $ x= x_F + x_G $

There must a property in the lesson that allows to state that, or maybe an intuitive way to see, but I am stuck at trying to find anything that could lead me to this conclusion. Can someone explain to the why there exists $x_F$ and $x_G$ such that $x_F + x_G = x \in F^{\perp}$?

Answer 1

There seems to be some error in the statement.

If $F$ and $G$ are orthogonal and $E=F\bigoplus G$, then any $x\in E$ can be written as $x=x_F + x_G$ with $x_F\in F$ and $x_G\in G$; in the particular case $x\in F^\perp$, then $x_F=0$ (the vector with all $0$ components) and $x_G=x$.

Note that if $x\notin E$, then $x$ can be written as $(x_F+x_G)+x_H$ with $x_H\in F^\perp\cap G^\perp$ and $x_F$, $x_G$ as above.