Field extension of a vector space If $V$ is a vector space over the field $k$, and $K$ is a field extension of $k$, then $(V)_K$ over $K$ is a vector space. How this new vector space is constructed? and how  are the linear transformations defined?
 A: This new space is constructed using the tensor product.  $V_K = K \otimes_k V$.  That means $V_K$ is spanned by symbols of the form $a \otimes v$ where $v \in V$ and $a \in k$.  The scalar multiplication by elements of $K$ is done by multiplying into the left factor
$$b(a \otimes v) = (ba) \otimes v.$$
When manipulating these symbols you are also allowed to cross elements of $k$ over the tensor symbol so
$$(ab) \otimes v = a \otimes bv$$
(but only when $b \in k$!!  If $b \in K \setminus k$ then it can't cross!).  And finally addition can come out of either factor
$$(a + b) \otimes v = a \otimes v + b \otimes v \qquad \text{and} \qquad a \otimes (v + w) = a \otimes v + b\otimes w.$$
Note that these rules are not enough to combine every sum into an element of the form $a \otimes v$.  Guys in that form are called simple tensors and in general you get all the elements of $V_K$ by taking sums of simple tensors, just the simple tensors alone doesn't cut it.
If $\{v_1, \ldots, v_n\}$ is a basis for $V$ as a $k$-vector space then $\{1 \otimes v_1, \ldots, 1 \otimes v_n\}$ is a basis for $V_K$ as a $K$-vector space.  Linear transformations of $V_K$ are defined just as they are for any vector space and are uniquely determined (as always) by where they send a basis.
