1

Fix a field extension $k\subseteq K$ and consider a linear system $Ax=b$ where $A$ is a matrix (not necessarily square) with coefficients in $k$. I don't understand why if the above linear system has a solution in $K$, then there exists also a solution in $k$.

Thanks in advance.

user26857
  • 52,094
Dubious
  • 13,350
  • 12
  • 53
  • 142
  • For square matrices this is a consequence of Cramer's rule. Maybe you can generalize this to your problem. –  May 16 '14 at 15:44
  • In my case, the matrix is not square. This is the point. – Dubious May 16 '14 at 15:46
  • 6
    The algorithm for finding the set of solutions to a linear system is immune to the existence of an extension field of $k$. For example the reduced row echelon form of $A$ remains the same. As does that of $(A|b)$. The claim follows from this. – Jyrki Lahtonen May 16 '14 at 15:58
  • The proposed duplicate focuses on how the real numbers extend the rationals, but Jyrki has identified the crux of the argument that applies in all field extension settings. Row reduction is our friend. – hardmath Feb 27 '16 at 16:43

1 Answers1

2

There exists a $k$-linear projection $\pi:K\to k$. Applying $(\pi,\ldots,\pi)$ to both sides of $Ax=b$, we see that if $(x_1,\ldots,x_n)$ is a solution, so is $(\pi(x_1),\ldots,\pi(x_n))$.

Andrew Dudzik
  • 30,074
  • This may lead to a trivial solution in $k,$ isn't it? – User Apr 11 '19 at 08:14
  • If $b=0$, then yes, it's possible. – Andrew Dudzik Apr 11 '19 at 08:18
  • So arbitrary projections cannot guarantee a solution in $k.$ – User Apr 11 '19 at 08:20
  • Why not? The trivial solution is still a solution. – Andrew Dudzik Apr 11 '19 at 13:54
  • If you have $b=0$ and you're trying to avoid the trivial solution, then if $x_jneq 0$ (for some $j$), you can choose a projection with $\pi(x_j) \neq 0$. (if $x_j\notin k$, pick a basis of $K/k$ that includes $1$ and $x_j - 1$) – Andrew Dudzik Apr 11 '19 at 14:03
  • Sorry for the late reply. I meant the statement, exactly, should be that the existences of non-trivial solutions in field extensions are equivalent, and arbitrary projection cannot serve the work. Of course the way how you picked a particular projection in the last comment is undoubytly correct! – User Apr 14 '19 at 13:45
  • @Slade how do know a projection $\pi$ exists? – Math_Day Dec 04 '22 at 20:52
  • @Math_Day This is a general fact about vector spaces: there exists a projection onto any subspace. If you choose a $k$-basis for $K$, any nonzero function on the basis extends to such a projection. – Andrew Dudzik Dec 05 '22 at 00:41