Is restriction of scalars well-defined on subspaces? Let $K/k$ be an extension of fields and let $v_1,\ldots,v_r,u_1,\ldots,u_r\in k^n$.  If the span of the $v$'s over $K$ equals the span of the $u$'s over $K$, must the two spans also be equal over $k$?
$$_K\langle v_1,\ldots,v_r\rangle=_K\langle u_1,\ldots,u_r\rangle\qquad\overset{?}\Longrightarrow\qquad_k\langle v_1,\ldots,v_r\rangle=_k\langle u_1,\ldots,u_r\rangle$$
Here is another way to look at this question: Given a subspace $U\subset K^n$ with some basis in $k^n$, is the procedure of restricting scalars to $k$ well-defined, or does it depend on the choice of basis of $U$?
 A: Step 1: A subset $S \subset k^n$ is $k$-linearly independent iff it is $K$-linearly independent.  This can be shown for instance by expressing linear independence in terms of the reduced row echelon form (rref) of the matrix with columns the elements of $S$ and appealing to the uniqueness of rref.
(Alternate proof: if $S$ is $k$-linearly independent, extend it to a $k$-basis $B$ of $k^n$.  The matrix $M$ with columns the elements of $B$ is invertible, i.e., there is a matrix $N$ with $MN = NM = 1$.   This matrix equation holds over $K$ as well and shows that the columns of $M$ are $K$-linearly independent.)  
Step 2: To ask whether 
$\langle u_1,\ldots,u_r \rangle_K \subset \langle v_1,\ldots,v_s \rangle_K$ 
we may assume without loss of generality that $v_1,\ldots,v_s$ are $k$-linearly independent, because by throwing away some of the $v_j$'s we can reduce to this case without changing the span (either over $k$ or over $K$).  Then the above containment holds iff for all $i$, the set $v_1,\ldots,v_s,u_i$ is $K$-linearly dependent.  As above, this holds over $k$ iff it holds over $K$, so the given containment implies
$\langle u_1,\ldots,u_r \rangle_k \subset \langle v_1,\ldots,v_s \rangle_k$.
Step 3: The affirmative answer to your question follows immediately from Step 2.
