Does a set polynomials (of all orders) of a complex variable form a linear vector space? Indeed all such polynomials would satisfy the commutative and associative property of addition (or so I think). However, one of the properties of vectors is that there must exist a unique null vector. But if vectors were indeed polynomials of say nth order, there should be n solutions to every polynomial when the vector becomes zero. So can we say it doesn't form an LVS because the null vector is remotely not unique?
 A: The zero vector is defined as the one that when added to any vector in the space does not change it.  In the vector space of polynomials, we do not care about roots.  The null vector is the zero vector, because that is the only one you can add to, say, $x$ and still get $x$.  You can verify all the axioms of a vector space and conclude that this is one.
Although we talk of polynomials, it may be better to view them as tuples of numbers.  The powers of $x$ are just there to keep track of the order of the numbers.  You are not supposed to find roots, take derivatives, factor, or any of the things we usually do with polynomials.  You can just add the tuples or multiply them by a scalar.
A: The polynomials of exactly degree $n$ do not form a vector space. Apart from the problem with the zero polynomial not being of degree $n$, we also have that the sum of two polynomials of equal degree may be a polynomial of lower degree. For example,
\begin{align}
p(x) &= 2x^2-3x+5 && \text{(degree 2)}\\
q(x) &= -2x^2+5x+3 && \text{(degree 2)}\\
\implies p(x)+q(x) &= 2x+8 && \text{(degree 1)}
\end{align}
However the polynomials of degree at most $n$ do form a vector space, of dimension $n+1$. Note that the zero polynomial typically gets assigned the formal degree $-\infty$, so it is included.
Also the set of all polynomials forms a vector space. This vector space has infinite dimension, as you need infinitely many basis vectors (for example, the polynomials of the form $x^k,k\in\mathbb N$ form a basis; omit just one of them, and there will be polynomials you no longer can form).
