A result on extension fields in linear algebra.

Question

Let $F$ be a subfield of $E$, $A$ an element of $\mathcal{M}_F(m,n)$ and $b$ a vector in $F^m \subset E^m$.

What is the easiest way to prove the following statement: if $Ax = b$ has a solution in $E^n$ then it has a solution in the subset $F^n$.

Answer 1

One may determine whether $Ax=b$ has a solution by using row operations. Since $A$ and $b$ have entries in $F$, this process is defined completely over $F$, without reference to the field $E$.

Answer 2

The fact that $Ax=b$ has a solution in $E^n$ tells you that the row rank of the augmented matrix $[A|b]$ is equal to the row rank of the matrix $A$, working with scalars in $E$. What you need to check (and it does follow from uniqueness of reduced echelon form, for example) is that $v_1,\dots, v_m\in F^n$ are linearly independent over $E$ if and only if they are linearly independent over $F$. So row rank working over $E$ is equal to row rank working over $F$.