Show that $X$, with the usual addition and the usual multiplication by real numbers, is a real vector space of dimension $n + 1$. Find a basis On a fixed interval $[a, b]\subset \mathbb{R}$, consider the set $X$ consisting of all polynomials with real coefficients and of degree not exceeding a given
$n$, and the polynomial $x = 0$ (for which a degree is not defined in the
usual discussion of degree). Show that $X$, with the usual addition and
the usual multiplication by real numbers, is a real vector space of
dimension $n + 1$. Find a basis for $X$. Show that we can obtain a
complex vector space $\bar{X}$ in a similar fashion if we let those coefficients
be complex. Is $X$ a subspace of $\bar{X}$?
I know that this is a real vector space, but I do not know how to prove that the dimension is $n+1$, I could say that $\{1, x, x^2, ..., x^n\}$ is a basis for this vector space? And with this it would be proven that it has dimension $n+1$?
What changes if instead of real the coefficients are complex and the scalars of the vector space also? Change the base? And the dimension?
$X$ is not a vector subspace of $\bar{X}$ since $2$ is a polynomial of degree $0$ in $\mathbb{R}$ and $i\in \mathbb{C}$ but $2i\notin\mathbb{R}$.
 A: You should have a theorem that says the number of elements in a basis is the dimension of the space.  Your proposed basis is a good one.  It has $n+1$ elements.  All you need to do is prove that it is a basis.  Nothing of what we have said changes if the field is $\Bbb C$ instead of $\Bbb R$.  Your argument that $X$ is not a vector subspace of $\overline X$ is correct.  A vector space is defined as a vector space over a field and a vector subspace has to be over the same field.
A: A linearly independent set which spans the entire space is a basis, by definition.
You know that $\{1, x, \ldots, x^{n+1}\}$ is linearly independent. It also spans the entire space, by definition:
$$p(x) = a_0 + a_1x + \cdots + a_nx^n \in \operatorname{span}\{1, x, \ldots, x^{n+1}\} $$
Hence, it is a basis.
As you noted, $X$ is not a subspace of $\overline{X}$ if the scalars are also complex.
But, the set of all complex polynomials of degree $\le n$ also forms a real vector space $\overline{X}_\mathbb{R}$.
The dimension of $\overline{X}_\mathbb{R}$ is $2n+2$ and one basis is:
$$\{1, x, \ldots, x^{n+1}, i, ix, \ldots, ix^{n+1}\}$$
It spans the whole space $\overline{X}_\mathbb{R}$:
\begin{align}
p(x) &= (a_0 + ib_0) + (a_1 + ib_1)x + \ldots (a_n + ib_n)x_n \\
&= a_0 + a_1x + \cdots + a_nx^n + b_0i + b_1ix + \cdots + b_nix^n
\end{align}
Remember, the scalars $a_0, \ldots, a_n, b_0, \ldots, b_n$ are real numbers.
It is also linearly indepepdent:
\begin{align}
0 &= a_0 + a_1x + \cdots + a_nx^n + b_0i + b_1ix + \cdots + a_nx^n \\
&= (a_0 + a_1x + \cdots + a_nx^n) + i(b_0 + b_1x + \cdots + b_nx^n)
\end{align}
The real and imaginary part must both separately be zero so we get that all scalars are zero.
And indeed $X = \operatorname{span}\{1, x, \ldots, x^n\}$ is now a subspace of $\overline{X}_\mathbb{R}$.
