Answer
Yes, this is a subspace of $\mathbb{P}_{n}$.
Work Step by Step
For
$ H=\{p\ \ | p(0)=0, \}$
following the definition of a subspace on p.195,
(a) Is the zero vector in $H$ ?
$p(x)=0$ is the zero vector, and
$p(0)=0$,
so, yes, the zero vector is in $H$
(b) Is H closed under addition?
If $\mathrm{p}$ and $\mathrm{q}$ are in $H$, then
$(\mathrm{p}+\mathrm{q})($0$)=\mathrm{p}(0)+\mathrm{q}(0)=0+0=0$,
so $\mathrm{p}+\mathrm{q}$ is in $H$.
Yes, H is closed under addition
(c) Is H closed under multiplication by scalars?
Let $p\in H, c\in \mathbb{R}$. Then
$(c\mathrm{p})(0)=c\cdot \mathrm{p}(0)=c\cdot 0=0$,
so $c\mathrm{p} \in H$.
H is closed under multiplication by scalars
So,
$H$ is a subspace of $\mathbb{P}_{n}$
