Quotient of pre-hilbert space

Question

Let $H$ be a pre-hilbert space and let $F$ be a complete subspace of $H$. Show that $H/F$ is a pre-hilbert space.

My attemp:

Consider the following theorem:

$\it{Theorem}$: Let $H$ a pre-hilbert space and let $F$ be a complete subspace of $H$, $F$ $\neq$ $\emptyset$ and $x \in H$. Then there exist an unique $f_{0} \in F$ such that $||x - f_{0}|| = dist(x, F)$. This $f_{0}$ is characterized by $x - f_{0} \in F^{\perp}$.

I defined $\phi: H/F \to F^{\perp}$ by $\phi(\bar{x}) = x - f_{0}$, where $f_{0}$ is given by the the above theorem.

I would like show that $\phi$ is linear isometry. I already prove that $||\phi(x)|| = ||x||$ and $\phi$ is sobrejetive, but i can't show that this mapping is linear.

Answer 1

Hint: if $x-f_0 \in F^{\perp}$ and $y-f_1 \in F^{\perp}$ with $f_0,f_1 \in F$ then $(ax+by)-(af_0+bf_1) \in F^{\perp}$ and $af_0+bf_1 \in F$ for any scalars $a$ and $b$. [I have used the elmenetary fact that $F^{\perp}$ is a linear space].