Is the zero vector always included in the orthogonal complement?

Question

This might be a really silly question, but I'm currently studying for my upcoming analysis exam, and I can't quite find an answer to this.

The reason I'm asking is actually, that I've been looking over sample questions for the exam, some of which are about determining whether a claim is true or false - the one that gave rise to this question is the following:

"Let $\mathcal{H}$ be a Hilbert space, and let $M$ be a closed subspace of $\mathcal{H}$. For each $x \in \mathcal{H}$ there is a $y\in M$ such that $x−y\in M^{\perp}$."

What I'm thinking is:

Assume that for all $x \in M$ $y \in M$ exists such that $x-y \in M^{\perp}$. That means, that $\langle x-y,m \rangle = 0$ which, by linearity in the first variable, means that:

$\langle x , m \rangle - \langle y , m \rangle = 0 \Leftrightarrow \langle x , m \rangle = \langle y , m \rangle \Rightarrow x=y \Rightarrow x-y=0 \in M^{\perp}$

Thus such a $y$ exists and is in fact $y=x$ - if the zero vector is indeed always contained in the orthogonal complement.

Is that even remotely true? I appologise if this is a silly question, but I'm having a hard time grasping this particular subject.

Any help would be much appreciated, thanks! :-)

Answer 1

Orthogonal complement must have $\vec{0}$ in it. After all, if $A$ is some set, $$A^\perp = \{ x | x\cdot a = 0\ \forall a \in A\},$$ and so $\vec{0} \in A^\perp$ by definition.

Answer 2

Yes, $0$ is always in the complement, just by definition.

However, you should show the existence of $y \in M$ for every $x \in \mathcal{H}$ and not just for $x \in M$.

Answer 3

"For each $x \in \mathcal{H}$ there is a $y\in M$ ..."

You showed something for each $x \in M$, but the exam question asked for each $x \in \mathcal{H}$, and (in general) not all elements of $\mathcal{H}$ are in $M$.

Answer 4

Yes it is included as $0$ is orthogonal to everything.

However if your question is more along the lines of "$0$ is a weird case so do we discriminate?".

We don't and for good reason! If we include $0$, the orthogonal compliment is a vector space so we have tons of theory already.