# A question on Isometry between the orthogonal subspaces of Hilbert spaces

I was reviewing my class-notes on Functional analysis and the professor had mentioned that given a closed proper subspace $U$ of an hilbert space $\mathcal{H}$, $\exists$ a closed subspace $U^{\perp}$ such that $U\oplus U^{\perp}=\mathcal{H}$ is an isometry from $\{<z,w>\} \to z+w$ where $z \in U,w\in U^{\perp}$

I know the definition of an isometry(that it is a distance preserving map between two metric spaces) but why is this map an isometry? And why isnt $U\oplus U^{\perp}$ the same as $\mathcal{H}$

I would be grateful if someone could help me out here. Thanks

There are two understandings of the notation $U \oplus V$.
The first interpretation means that if you write $U \oplus V$, then $U$ and $V$ are both subspaces of a common inner product space and orthogonal to each other (i.e. $\langle x,y\rangle =0$ for $x\in U,y\in V$ and $U \oplus V=\{x+y\mid x\in U, y \in V\}$. In this interpretation, you have $\mathcal{H}=U \oplus U^\bot$.
In the second interpretation, $U,V$ can be different spaces(i.e. they don't have to be subspaces of a common inner product space) and
$$U \oplus V :=\{ (u,v)\mid u\in U, v\in V\},$$
equipped with the usual vector space structure and with the scalar product $$\langle (u,v), (a,b)\rangle = \langle u,a\rangle +\langle v,b \rangle.$$ This is the interpretation used in your statement.
That your map is an isometry is a consequence of Pythagoras theorem, ie that $\Vert u+v\Vert^2= \Vert u\Vert^2 +\Vert v\Vert^2$ if $u,v$ are orthogonal.