Prove a map is linear ... or not Let F[x] denote the vector space of polynomials in the variable x with coefﬁcients in F. Let e1 : F[x] -> F denote the map taking a polynomial f to f(1) $\in$ F. Prove or disprove e1 is linear.
To begin with I have no idea what F[x] is. What does it mea to be a vector space of polynomials IN the variable x?
 A: Hint:
$F[x]$ is the set of all polynomials in one variable, named $x$, with coefficients in the field $F$:
$$
F[x]=\{a_0+a_1x+a_2x^2+\cdots+a_nx^n\mid a_0,\ldots,a_n\in F,\ n\in\mathbb{N}\}.
$$
Note that we can view $F[x]$ as a vector space over $F$ by defining scalar multiplication and vector addition. If $\alpha\in F$ is a scalar, and $f,g\in F[x]$ are the polynomials
$$\tag{1}
f(x)=b_0+b_1x+\cdots+b_nx^n,\qquad g(x)=c_0+c_1x+\cdots+c_nx^n
$$
(where the polynomial of lower degree will have zeroes for its highest coefficients), then we define $\alpha\cdot f$ to be the polynomial
$$\tag{2}
(\alpha\cdot f)(x):=\alpha\cdot f(x)=\alpha b_0+\alpha b_1x+\cdots+\alpha b_nx^n
$$
and we define $f+g$ to be the polynomial
$$\tag{3}
(f+g)(x):=f(x)+g(x)=(b_0+c_0)+(b_1+c_1)x+\cdots+(b_n+c_n)x^n.
$$
Now: given $F[x]$, there is a family of natural functions $\phi_a:F[x]\rightarrow F$ for $a\in F$ defined by
$$
\phi_a(f):=f(a)=b_0+b_1a+b_2a^2+\cdots+b_na^n.
$$
(These are called the evaluation maps, I believe.) This problem asks about $\phi_1$ in particular, where $1$ is the multiplicative identity in the field $F$. How does $\phi_1$ behave? Using the field properties, we see
$$\tag{4}
\phi_1(f)=f(1)=b_0+b_1\cdot1+b_2\cdot1^2+\cdots+b_n\cdot 1^n=b_0+b_1+\cdots+b_n.
$$
So, the question here is this: is $\phi_1$ a linear mapping? In other words, for $f$ and $g$ as written in (1), $\alpha\cdot f$ and $f+g$ as defined in (2) and (3) respectively, and $\phi_1$ the mapping from (4) that takes a polynomial and gives back the sum of its coefficients, is it true that
$$
\phi_1(\alpha\cdot f)=\alpha\cdot\phi_1(f)\qquad\text{and}\qquad \phi_1(f+g)=\phi_1(f)+\phi_1(g)
$$
holds?
