Is it a subspace?

Definition: Let $V$ be a vector space over $F$ and let $A \in \text{End}(V)$, with $A$ nilpotent operator with index $r$. The height of a vector $v \in V$ relative to the operator A, it is the minimum positive integer k for which it is satisfied that $A^{k} (v) = 0$.

This is denoted $h(v)=k.$

Let's notice that if

$$A^{r}(u)=0\ \ \forall u \in V,$$

so

$$h(u) \leq r\ \ \forall u\in V.$$

If $A^{r-1} \neq 0$ then there are vectors $w \in V$ such that $h(w)=r$.

Let $V_{s}=\{v \in V: h(v) \leq s\}$ for $s \in \{0,1,...,r\}$.

Is $V_{s}$ a subspace?.

I know that $0\in V_{s}$ for $0 \leq s \leq r$. But I don't know how to prove that if $u, v \in V_{s}$ then $u+v \in V_{s}$ and if $ \alpha \in F$ and $v \in V_{s}$ then $\alpha v \in V_{s}$.

I really appreciate any help.

$A^{i}u=0$ and $A^{j}v=0$ imply $A^{k} (\alpha u+\beta v)=0$ where $k=\max \{i,j\}$.

[Note that $A^{i}u=0$ implies $A^{m}u=0$ for all $m \geq i$. So $A^{k}(\alpha u+\beta v)=\alpha A^{k}u+\beta A^{k}v=0$]