Let $L$ and $K$ be fields with $L \subset K$. Let $v_1,\ldots,v_r \in L^n$ be column vectors, linearly independent over $L$. Of course, we can also consider the vectors to sit in $K^n \supset L^n$.
Question: Is there a simple explanation, not involving determinants, for why $v_1,\ldots,v_r$ are also linearly independent over the larger field $K$?
Notes:
- When $n=r$, this follows by considering the $\det ( v_1 \cdots v_n)$, which is a number in $L$.
- When $n > r$, you can probably do something similar using subdeterminants.
- Sometimes you have nice automorphisms which make this easy: e.g. if $v_1,\ldots,v_r \in \mathbb{R}^n$ are linearly independent and $z_1,\ldots,z_n \in \mathbb{C}$ have $z_1 v_1+\ldots z_n v_n =0$, then you can use the formulas $\frac{z+\overline z}{2}, \frac{z - \overline z}{2 i}$ for the real and imaginary parts, as well as the fact that $\mathbb{R}$ is fixed by conjugation to conclude the real/imaginary parts of all the $z_i$ are zero.
But how do we do this in general, without using determinants? Maybe some argument involving polynomials?