Is the $0$ vector of a linear subspace the same as the $0$ of the vector space? I'm asking this because I'm trying to prove that $P_s(\mathbb R)$ is a linear subspace of $P_n(\mathbb R)$, where
$$P_n(\mathbb R) = a_0x^0 + a_1x^1+\cdots + a_nx^n$$
If $P_s(\mathbb R)$ is a linear subspace of $P_n(\mathbb R)$, then $P_s(\mathbb R)$ should contain the $0$ vector of $P_n(\mathbb R)$, which is 
$$(0,0,\cdots,0)\tag{$n$ times}$$
but this isn't a zero vector of $P_s(\mathbb R)$, which would be 
$$(0,0,\cdots,0)\tag{$s$ times}$$
So how should I fix it?
 A: There are some distinctions to make here; you can only say, for instance, that $\mathbb{R}^1$ - that is the set of real $1$-tuples - is a subspace of $\mathbb{R}^2$ - the set of pairs - in the sense that there is a subspace of $\mathbb{R}^2$ isomorphic to $\mathbb{R}^1$ - more precisely, there is a linear surjection $\mathbb{R}^2\rightarrow\mathbb{R}$. However, as you note, since elements of $\mathbb{R}^1$ are not pairs, it is not even a subset of $\mathbb{R}^2$ and hence not really a subspace.
In the case of polynomials, however, $P_s(\mathbb R)$ is actually a subspace of $P_n(\mathbb R)$ - the zero vector is both spaces is not a tuple, but the $0$ polynomial. What you are seeing is that there is a canonical map $f_n:P_n(\mathbb R)\rightarrow \mathbb R^n$ defined as
$$a_0x^0+a_1x^1+\ldots + a_nx^n\mapsto (a_0,a_1,\ldots,a_n)$$
and that the zero of $f_n(P_n(\mathbb R))$ is not the same as in $f_s(P_s(\mathbb R))$ - however, why should it be? $f_n$ and $f_s$ have different domains. To clarify further, note that $f_n(P_s(\mathbb R))$ is well-defined and it maps to $n$-tuples (with $n-s$ zeros at the end), and has the same zero element as desired.
A: This is basically a matter of notation. If you think about the zero vectors as polynomials, you will notice that it is in fact the same polynomial, i.e. the same element of $P_n(\mathbb{R})$, in both cases. The reason they have different numbers of $0$s is just because you have chosen different ways of writing elements of $P_n(\mathbb{R})$ and $P_s(\mathbb{R})$ in terms of coordinates.
