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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?

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    $\begingroup$ I think that a simple Gauss-Jordan elimination argument works. $\endgroup$ – Quang Hoang Oct 19 '15 at 15:17
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    $\begingroup$ One idea: If we have an $L$-linear map $f:K\to L$, then $f(k_1 v_1 + \cdots + k_n v_n) = f(k_1) f(v_1) + \cdots + f(k_n)f(v_n) = f(k_1) v_1 + \cdots + f(k_n)v_n = 0$, implying that all $f(k_i) = 0$. If $K/L$ is finite, we can take the trace in characteristic $0$ and something like the Frobenius map in characteristic $p\not = 0$. Use a Zorn's lemma argument to handle the general case. $\endgroup$ – anomaly Oct 19 '15 at 15:23
  • $\begingroup$ Thanks, I'll investigate these ideas later on today. $\endgroup$ – Mike F Oct 19 '15 at 18:56
  • $\begingroup$ I want to give another viewpoint. Suppose $f\colon L^r\to L^n$, $e_j\mapsto v_j$. By linear independence, $f$ is injective. Note that $F^n=L^n\otimes_LF$ and the same for $F^r$, and the functor $\otimes_LF$ is exact, since $F$ is a free (hence flat) $L$-module. Hence, $f\otimes1\colon F^r\to F^n$ is injective. For reference of these terminologies, see Atiyah and Macdonald, and any reference for homological algebra. $\endgroup$ – Yai0Phah Nov 13 '15 at 12:39

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