Condition for Linear Dependence of Linear Functionals in terms of Kernel

Theorem
Let $V$ be a Definition:Vector Space over a field $\GF$.

Let $f, f_1, \ldots, f_n: V \to \GF$ be linear functionals.

Suppose that:


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

where $\ker f$ denotes the kernel of $f$.

Then there exist $\alpha_1, \ldots, \alpha_n \in \GF$ such that:


 * $\ds \forall v \in V: \map f v = \sum_{i \mathop = 1}^n \alpha_i \map {f_i} v$

Proof
For $i = 1, \ldots, n$, let $w_i$ be such that:


 * $w_i \not\in \ker f_i$
 * $w_i \in \ker f_j, j \ne i$

Suppose first that the $w_i$ all exist.

Let $v_i = \dfrac 1 {\map {f_i} {w_i} } w_i$.

Then since $f_i$ is linear:


 * $\map {f_i} {v_i} = 1$

Furthermore for $j \ne i$, $\map {f_j} {v_i} = 0$.

Now let $v \in V$ be arbitrary, and define:


 * $\ds w = v - \sum_{i \mathop = 1}^n \map {f_i} v v_i$

Then:

Since $j$ was arbitrary, it follows from the premise on $f$ that:


 * $\map f w = 0$

That is to say:


 * $\ds 0 = \map f v - \sum_{i \mathop = 1}^n \map {f_i} v \map f {v_i}$

Setting $\alpha_i = \map f {v_i}$ we find:


 * $\ds \map f v = \sum_{i \mathop = 1}^n \alpha_i \map {f_i} v$

Now the case that such $w_i$ does not exist for a particular $i$.

Then:


 * $\ds \bigcap_{i \mathop = 1}^n \ker f_i = \bigcap_{j \mathop \ne i} \ker f_j$

and $\set{ f_j: j \ne i}$ fulfils the condition of the result.

The result for $f_1, \ldots, f_n$ follows by setting $\alpha_i = 0$.