Halmos Finite-Dimensional Vector Spaces: Does $\mathcal{P}$ over $\mathbb{C}$ with $x(t) = x(1 - t)$ form a vector space? Paul R. Halmos "Finite-Dimensional Vector Spaces", 2e, chapter I, section 2, exercise 5.d:

Consider the vector space $\mathcal{P}$ and the subsets $\mathcal{V}$
  of $\mathcal{P}$ consisting of those vectors (polynomials) $x$ for
  which
(d) $x(t) = x(1 - t)$ for all $t$.
In which of these cases is $\mathcal{V}$ a vector space?

I would suggest that the subset satisfying (d) does form a vector space, since 


*

*(d) forms a linear constraint in the polynomial's coefficients $\mathbf{a}$ as in


$$
g(\mathbf{a}) = x(t) - x(1 - t) = 0,
$$


*

*the zero vector is included,

*every element has an inverse element.


$g$ is a linear constraint, since 
$$
g(\mathbf{a} + \mathbf{b}) = g(\mathbf{a}) + g(\mathbf{b}) \qquad \wedge \qquad \alpha g(\mathbf{a}) = g(\alpha \mathbf{a}),
$$
where $\mathbf{a}$, $\mathbf{b}$ are coefficients of polynomials in $\mathcal{P}$, and $\alpha$ is a complex number.
Is that correct?
 A: 
I would suggest that the subset satisfying (d) does form a vector space, since 
  
  
*
  
*(d) forms a linear constraint in the polynomial's coefficients,
  
*the zero vector is included,
  
*every element has an inverse element.
  
  
  Is that correct?

This is the right idea, but when you say that (d) forms a linear constraint in the coefficients, what do you mean?
I suppose what you mean is that if the polynomial is $a_0 + a_1 x + a_2 x^2 + \cdots ...$ then it is some linear equation in $a_0, a_1, \ldots$. But why does this imply that (d) is a vector space?
However you can formalize (d). I would approach this by defining a map
$$
A : \mathcal{P} \to \mathcal{P}
$$
where $A(\boldsymbol{x}) = \boldsymbol{x}(t) - \boldsymbol{x}(1-t)$.
Then, show that $A$ is a linear map -- that is, it preserves addition and scalar multiplication.
Finally, $\mathcal{V}$ is the set of polynomials $\boldsymbol{x}$ such that $A(\boldsymbol{x}) = \boldsymbol{0}$. Since $A$ is linear this is a vector subspace. So this completes your proof.
A: Your first bullet point is not useful. Just verify the definition. Surely 0 is in there. Every element has an additive inverse. If $c$ is a constant, then $cx(t)=cx(1-t)$, so constant multiples of polynomials are also in there. So you're done.
