System of linear equations

System of linear equations

For $A \in \mathbf{F}^{m\times n}$ and one particular $b \in \mathbf{F}^n$, then

$Ax = b$ has exactly one solution $\iff$ ($ A $ injective) and ($ b \in \operatorname{range} A $).

Suppose $ T \in \mathcal{L}(V,W) $ and $ c \in W $, then

$ {x \in V : Tx = c} $ is either the empty set or is a translate of $ \operatorname{null} T $.

Special case: system of linear equations

general solution = particular solution + homogeneous solution

$ V/\operatorname{null} T \cong_\tilde{T} \operatorname{range} T $

If a linear system with all coefficients and constants in $\mathbb{Q}$ has a solution in $\mathbb{F} \supseteq \mathbb{Q}$, it has a solution in $\mathbb{Q}$.

Row-reduce the augmented matrix. Gaussian elimination uses only addition, subtraction, multiplication, and division by nonzero pivots — so starting from rational entries, every intermediate entry stays rational, and the resulting echelon form $R$ is rational.

The very same row operations are legitimate over $\mathbb{F}$ and produce the very same $R$. Now, if the system had no solution over $\mathbb{Q}$, the reduction would exhibit a row of the form $(0\;0\;\cdots\;0 \mid c)$ with $c \neq 0$ — but that row is equally an obstruction over $\mathbb{F}$, contradicting the existence of a solution there. So the reduction is consistent, and back-substitution (again only rational arithmetic) produces a solution with all $x_k \in \mathbb{Q}$. $\blacksquare$

Field-independent (determined over the small field $\mathbb{K}$): the minimal polynomial. Whether you compute it over $\mathbb{K}$ or over a bigger $\mathbb{F}$, you get the identical polynomial.

Field-dependent (can change when you enlarge): eigenvalues, eigenvectors, diagonalizability. These are about roots of polynomials, and roots can require a bigger field to exist.