Polynomial vector space terminology Consider the vector space $P$ and the subset $V$ of $P$ consisting of those vectors (polynomials) $x$ for which
a) $2x(0) = x(1)$,
b) $x(t) = x (1-t)$ for all $t$.
In which of these cases is $V$ a vector space?
What do $x(0)$ and $x(1)$ mean? 
Please help.
 A: To prove that $V$ is a vector subspace of $P$, one has to justify two things:


*

*If $x=x(t)$ and $y=y(t)$ are in $V$ then $x-y$ is in $V$

*If $a$ is a scalar up on the polynomial are taken, then $ax$ it is also in $V$.


For the first: if $x,y$ are in $V$ then both comply 
$$2x(0)=x(1)\quad \mbox{and}\quad 2y(0)=y(1)$$
so 
$$2(x-y)(0)=2x(0)-2y(0)=x(1)-y(1)=(x-y)(1).$$
Also, knowing that
$$x(t)=x(1-t)\quad \mbox{and}\quad y(t)=y(1-t)$$ 
then
$$(x-y)(t)=x(t)-y(t)=x(1-t)-y(1-t)=(x-y)(1-t),$$
hence $x-y\in V$.
For the second,
$$2ax(0)=a2x(0)=ax(1),$$
and
$$ax(t)=ax(1-t),$$
so $ax\in V$. 
A: I sympathise with your confusion at writing $x(1)$ for the result of evaluating the polynomial $x\in K[t]$ at $t=1$; out of dislike of this notation that uses polynomials as if they were functions (which they are not) I personally write $x[t:=1]$, or $x[1]$ for short. This notation can also be used for substituting a polynomial (like $1-t$) for $t$ into $x$, giving $x[t:=1-t]$, or $x[1-t]$ for short. Note that the subtle change in notation also avoids ambiguity: in b) to the question it would be quite natural to interpret $x(1-t)$ as a product rather than (as intended) as a substitution. It should be noted also that $x$ is a rather unusual since potentially confusing name for designating an arbitrary polynomial; I'll stick to the more common $p,q$.
Anyway, the main thing to know here is that for any $q\in K[t]$, the map $K[t]\to K[t]$ given by $p\mapsto p[q]$ is a linear map (in particular this holds when $q$ is a constant polynomial, in which case $p[q]$ is constant as well); this is straightforward to check. Now


*

*The set of a) in the question is the kernel of the map $K[t]\to K\subset K[t]$ given by $p\mapsto2p[0]-p[1]$;

*The set of b) in the question is the kernel of the map $K[t]\to K[t]$ given by $ p\mapsto p-p[1-t]$.

