Wrong argument? Matrix multiplication Let $f:K \rightarrow K$ be a linear transformation defined as $f(x) = a\cdot x$ where $a\in K$ and $K$ is a field extension of $F$. Let $A$ be the matrix that represents the transformation hence we obtain (where $I$ is the identity matrix):
$Ak = a\cdot k = a\cdot Ik$ for all $k\in K$ and thus we conclude $A = a\cdot I$.
Is this true? I think it works but I have tried this argument with an example and the equality I end up with was strange. 
 A: The key here is that $A$ is most likely meant to represent a linear transformation of $K$ as a vector space over $F$.  The matrix representing a linear transformation over $F$ should have entries that lie in $F$, it cannot have entries that lie in $K \setminus F$.
Maybe an example would help.  Let $K = \mathbb C$, which is an extension of $F = \mathbb R$ and let $a = i$ (the square root of $-1$).  As the base field is $\mathbb R$ the entries of our matrix must be real numbers, so $i$ cannot be in our matrix.  To get the matrix observe that $\mathbb C$ has basis $1, i$ over $\mathbb R$.  Mult by $i$ sends $1 \mapsto i$ and $i \mapsto -1$ so we get
$$A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$$
The characteristic of $A$ is $x^2 + 1$ and $i$ is indeed a root.
On the other hand, if we think of $\mathbb C$ as a vector space over itself then the entries of our matrix are allowed to lie in $\mathbb C$.  To get the matrix observe that $\mathbb C$ has basis $1$ and mult by $i$ sends $1 \mapsto i\cdot 1$ so
$$A = \begin{bmatrix} i \end{bmatrix}$$
The characteristic is $x - i$ and $i$ is a root, but in this case the result is trivial so it's much more likely that you are supposed to consider the map as linear over $F$ and not linear over $K$.
