Condition for Linear Dependence of Linear Functionals in terms of Kernel/Proof 2

Proof
Define $T : X \to {\GF}^n$ by:


 * $\map T x = \paren {\map {g_1} x, \map {g_2} x, \ldots, \map {g_n} x}$

for each $x \in X$.

We show that $T$ is linear.

Let $x, y \in X$ and $\alpha, \beta \in \GF$.

We have:

so $T$ is linear.

Now, we have $\map T x = 0$ $\map {g_i} x = 0$ for each $x \in X$.

That is, :


 * $\ds x \in \bigcap_{i \mathop = 1}^n \ker g_i$

so that:


 * $\ds \ker T = \bigcap_{i \mathop = 1}^n \ker g_i \subseteq \ker f$

Now, suppose that $x, y \in X$ have $T x = T y$.

Then by the linearity of $T$, we have:


 * $\map T {x - y} = 0$

so $x - y \subseteq \ker f$.

So $\map f {x - y} = 0$, and hence $\map f x = \map f y$, for these $x, y \in X$.

We can therefore define a map $\widetilde f : \Img T \to \GF$ by:


 * $\map {\widetilde f} {\map T x} = \map f x$

for each $x \in \Img T$.

From Image of Vector Subspace under Linear Transformation is Vector Subspace, $\Img T$ is a vector subspace of ${\GF}^n$.

Let $\set {e_1, \ldots, e_n}$ be the standard basis for ${\GF}^n$.

Let $\set {v_1, \ldots, v_k}$ be a basis of $\Img T$.

From Linearly Independent Set is Contained in some Basis, $\set {v_1, \ldots, v_k}$ is contained in some basis $\set {v_1, \ldots, v_n}$ of ${\GF}^n$.

Define the mapping $h : {\GF}^n \to \GF$ by taking:


 * $\ds \map h {v_i} = \map {\widetilde f} {v_i}$ for $1 \le i \le k$

and:


 * $\ds \map h {v_i} = 0$ for $k < i \le n$

and taking:


 * $\ds \map h {\sum_{i \mathop = 1}^n \alpha_i v_i} = \sum_{i \mathop = 1}^n \alpha_i \map h {v_i}$

for scalars $\alpha_1, \alpha_2, \ldots, \alpha_n \in \GF$.

By construction, $h$ is a linear functional.

We then have, for each $x \in X$:

with $\map h {e_i} \in \GF$.

So we have:


 * $f \in \span \set {g_1, \ldots, g_n}$