Show that it is a K-linear map Let $K \leq K(a)$ a field extension with $[K(a):K]=n$.
$K(a)$ is a vector space over $K$.
How can I show that a map $\varphi : K(a) \rightarrow K(a)$, $\varphi(e)=ae$, is a K-linear map??
 A: If $\Bbb F$ is any field, and $a \in \Bbb F$ is any element of $\Bbb F$, then the map $\phi_a:\Bbb F\ \to \Bbb F$, $\phi_a(e) = ae$ for all $e \in \Bbb F$ is $\Bbb F$-linear, as follows from the basic definition of a linear map, to wit:
1.)  $u, v \in \Bbb F \Rightarrow \phi_a(u + v) = a(u + v) = au + av = \phi_a(u) + \phi_a(v);$
2.)  $v, \beta \in \Bbb F \Rightarrow \phi_a(\beta v) = a \beta v = \beta av = \beta \phi_a(v)$.
(1) and (2) show that $\phi_a$ satisfies the definition of $\Bbb F$-linearity;  it follows that if $\Bbb K \subset \Bbb F$ is a subfield, the map $\phi_a:\Bbb F \to \Bbb F$ is $\Bbb K$-linear.  Taking $\Bbb F = \Bbb K(a)$ then resolves the question at hand.
Note:  I think it should be pointed out, in the light of Timbuc's comment, we need assume $\phi(k) = k$ for $k \in \Bbb K$ if $\phi$ is to be a field automorphism of $\Bbb F \supset \Bbb K$ as well as being $\Bbb K$-linear; for then, since $\phi(k u) = \phi(k)\phi(u)$, $\Bbb K$-linearity follows from $\Bbb K$ being a fixed field of $\phi$.  But if $\phi$ is to be merely linear, as are the $\phi_a$ defined above, we needn't assume $\Bbb K$ fixed by $\phi$.  End of Note.
Note Added Friday 24 October 2014 12:05 AM PST:  This in response to the comment of Mary Starr (see below):  To find the characteristic polynomial of a linear map such as $\phi_a$, it helps to have a convenient basis of $K(a)$ in which to express $\phi_a$.  To this end we observe that, since $[K(a):K] = n$, $a$ satisfies a unique monic irreducible polynomial $f(x) \in K[x]$ with $\deg f(x) = n$; furthermore, the set $\mathcal A = \{1, a, a^2, \ldots, a^{n - 1} \} = \{a^j \mid 0 \le j \le n - 1 \}$ is linearly independent over $K$; thus, since $\dim K(a)$ over $K$ is $n$, $\mathcal A$ forms a basis over $K$ for $K(a)$; the action of $\phi_a$ on this basis is easily calculated.  We have, for $0 \le j \le n -2$, $\phi_a(a^j) = aa^j = a^{j + 1}$; to calculate $\phi_a(a^{n - 1})$ we write
$f(x) = \sum_0^n f_i x^i$, where the $f_i \in K$ and $f_n = 1$; then $a^n = -\sum_0^{n - 1} f_i a^i$, whence $\phi_a(a^{n - 1}) = aa^{n - 1} = a^n =  -\sum_0^{n - 1} f_i a^i$.  If we now identify the elements of $\mathcal A$ with the standard basis vectors of $K^n$ in the obvious fashion, so that
$1 \leftrightarrow \begin{pmatrix} 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}, \tag{1}$
$a \leftrightarrow \begin{pmatrix} 0 \\ 1 \\ \vdots \\ 0 \end{pmatrix}, \tag{2}$
right on down to
$a^{n - 1} \leftrightarrow \begin{pmatrix} 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix}; \tag{3}$
it is easy to see that the matrix of $\phi_a$ in this basis takes the form
$[\phi_a] = \begin{bmatrix} 0 & 0 & 0 & \ldots & 0 &  -f_0 \\ 1 & 0 & 0 & \ldots & 0 & - f_1 \\
0 & 1 & 0 & 0 & \ldots & -f_2 \\ \vdots \\ 0 & 0 &  \ldots & 1 & 0 & -f_{n - 2} \\
0 & 0 & \ldots & 0 & 1 & -f_{n - 1} \end{bmatrix}. \tag{4}$
We can from (4) calculate the characteristic polynomial $p_{\phi_a}(x)$ of $\phi_a$, i.e. determine
$p_{\phi_a}(x) = \det [\phi_a - x I] = \det \begin{bmatrix}  -x & 0 & 0 & \ldots & 0 &  -f_0 \\ 1 & -x & 0 & \ldots & 0 & - f_1 \\
0 & 1 & -x & 0 & \ldots & -f_2 \\ \vdots \\ 0 & 0 &  \ldots & 1 & -x & -f_{n - 2} \\
0 & 0 & \ldots & 0 & 1 & -f_{n - 1} -x \end{bmatrix}; \tag{5}$
since I am not expert at evaluating determinants, at this point I am going to point to a citing which affirms that in fact $p_{\phi_a}$ is in fact $f(x)$; I suspect this well-known result may be had by a not-too-difficult induction, but since I haven't yet worked that out, I will for the moment have to be content with the efforts of others in validating this assertion.  In any event, these comments show how and why the characteristic polynomial of $\phi_a$ is $f(x)$.
As a final comment, I think that, in the light of the above discussion of the original problem, it is worth pointing out that we have just shown that the characteristic polynomial of $\phi_a(e) = ae$ is $f(x) \in \Bbb K[x]$, when $\phi_a$ is construed as linear mapping of $\Bbb K(a)$ over the field $\Bbb K$.  When considered as a linear mapping over the field $K(a)$, $p_{\phi_a}(x) = a - x$, since then we have
$[\phi_a] = [a] \tag{6}$
for the matrix of $\phi_a$, since $K(a)$ is one-dimensional over itself.
In either case, we have $p_{\phi_a}(a) = 0$.  End of Note.
Hope this helps.  Cheerio,
and as ever,
Fiat Lux!!!
