Hoffman and Kunze: Linear Algebra section 1.4 remark How to prove the following result:
Given $AX = Y$. Suppose the entries of the matrix $A$ and the scalars $y_1, \dots, y_m $ lie in a subfield $F_1$ of the field $F$.
If $AX = Y$ has a solution with $x_1, \dots, x_n$ in $F$, then it has solution $x_1, \dots, x_n$ in $F_1$.
It mentions over either fields, the condition for the system to have a solution is that certain relations hold between $y_1, \dots, y_m$ in $F_1$.
Does that mean that given one can add more rows below matrix $A$, let’s say matrix $A'$. If matrix $A'$ has a solution in field $F$, then obviously $A$ has a solution in the subfield $F'$?
 A: Let $\pi:F\to F_1$ be an $F_1$-linear map which restricts to the identity on $F_1$. There is an induced map on the vector spaces $\pi_n:F^n\to F_1^n$, $(x_1,...,x_n)\mapsto (\pi(x_1),...,\pi(x_n))$. Then $A\pi_n(X)=\pi_m(AX)$ by $F_1$-linearity as $A\in M_{n\times m}(F_1)$. Since $Y\in F_1^m$, it is fixed by $\pi$, so $A\pi_n(X)=\pi_m(Y)=Y$ and we get a solution $\pi_n(X)\in F_1^n$.
To see that such maps exist, extend $\{1\}$ to a basis $B$ for $F$ (over $F_1$). Choose $\pi:F\to F_1$ sending $b\mapsto 0$ for $b\in B\setminus\{1\}$ and $1\mapsto 1$.
A: This is actually a rather subtle notion, which will require some content later in the book to nail down.
$A$ is an $m\times n$ matrix with entries in $F_1$. Consider two vector spaces: (1) $V_1$, the $F_1$-column space of $A$ — the subspace of $F_1^m$ consisting of all $F_1$-linear combinations of the columns, and (2) $V$, the $F$-column space of $A$ — the subspace of $F^m$ consisting of all $F$-linear combinations of the columns. Then the key point is that the dimension over $F_1$ of $V_1$ will equal the dimension over $F$ of $V$. Indeed, $V_1$ will be the intersection of the $V$ with $F_1^n$. We deduce that if $Y\in F_1^n$ is an $F$-linear combination of the columns of $A$, then it is in fact an $F_1$-linear combination of the columns of $A$.
We claim that if $a_1,\dots,a_k\in F_1^m$ are linearly independent columns of $A$, then they are likewise linearly independent in $F^m$. If $F$ is a finite-dimensional vector space over $F_1$, we can do it directly from the definition of linear independence. But, in general, probably the easiest approach is to use determinants. Consider the $m\times k$ matrix whose columns are the vectors $a_1,\dots, a_k$. They are linearly independent if and only if some $k\times k$ minor is nonzero, and that nonzero minor doesn't change when we view the vectors as being in $F^m$.
A: We have $Ax=y$ with entries in $A$ and $y$ coming from a subfield $F'$ of some field $F$.
How do we go about solving it?
We apply row operations to reduce $A$ to its row reduced echelon form $R$; the same operations change $y$ to some $y'$. We get the following (row equivalent- this means that the set of solutions of this system is same as that of the given system) system of equations:
$$Rx=y'\tag 1$$
If we regard entries of $R$ and those of $y$ to have come from $F$, then $(1)$ has a solution, whence it follows that rank $R$ is same as that of rank of the augmented matrix $[R|y']$. But this is same as saying that the system $Rx=y'$ is consistent in $F'$.
A: $\underline{\text{Claim:}}$
For a field $F$:
$\forall \color{red}{F_1}\leq \color{blue}{F},\text{ }\forall A\in M_{m\times n}(\color{red}{F_1}),\text{ }\forall Y\in \color{red}{F_1}^m:$

$$\bigg[(\exists \widetilde{X}\in \color{blue}{F}^n:A\widetilde{X}=Y)
\implies (\exists X\in \color{red}{F_1}^n:AX=Y)\Bigg].$$

$\underline{\text{Pf:}}$
Take $A\widetilde{X} = Y$ and consider the $F_1$-reduced row-echelon form system:
$$A_{rref}\widetilde{X}= Y'$$
$$\begin{bmatrix}I_{r\times r} && 0_{r\times(n-r)}\\ 0_{(m-r)\times r}&& 0_{(m-r)\times (n-r)} \end{bmatrix}_{m\times n}\begin{bmatrix}\widetilde{X}_1\\ \vdots \\ \widetilde{X}_n\end{bmatrix} = \begin{bmatrix}Y'_1\\ \vdots \\ Y'_m\end{bmatrix}$$
Whereby $F_1$-reduction, I mean replace $Row_j$ with $(\lambda Row_i + Row_j)$ using only $\lambda\in F_1$.

Converting back to equation form gives:
$$\begin{matrix}\widetilde{X}_1 = Y'_1 \\ \vdots \\ \widetilde{X}_r = Y'_r\\ \text{ }\\ 0 = Y'_{r+1} \\ \vdots \\ 0=Y'_m\end{matrix}$$
So, depending on the rank of $A$, we have $n-r$ free variables $\{\widetilde{X}_{r+1},...,\widetilde{X}_n\}$.
We may thus introduce new equations:
$$\begin{matrix}\widetilde{X}_{r+1} = t_{r+1} \\ \vdots \\ \widetilde{X}_n = t_n\end{matrix}$$
for choice of constants $t_i\in F$, which we may take to all be in $F_1$!
Also, since we know $A\widetilde{X}=Y$ has a solution, we don't get a contradiction with the latter rows $0 = Y'_i$ for $i\in \{r+1,...,m\}$

To complete the proof, define:
$$X:= \begin{bmatrix}Y'_1\\ \vdots \\ Y'_r \\ t_{r+1} \\ \vdots \\t_n\end{bmatrix}\in F_1^n.$$
By design, we have:
$$A_{rref}X = Y'$$
Assuming we kept track of the elementary row matrices (which are invertible), we can un-reduce the system to get: $AX = Y$ and we're done. $\blacksquare$
