2
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.