Condition for Linear Dependence of Linear Functionals in terms of Kernel/Proof 1
Theorem
Let $V$ be a 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:
- $\ds \forall v \in V: \map f v = \sum_{i \mathop = 1}^n \alpha_i \map {f_i} v$
That is:
- $f \in \span \set {f_1, \ldots, f_n}$
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:
\(\ds \map {f_j} w\) | \(=\) | \(\ds \map {f_j} v - \sum_{i \mathop = 1}^n \map {f_i} v \map {f_j} {v_i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {f_j} v - \map {f_j} v\) | $\map {f_i} {v_i} = 1, \map {f_j} {v_i} = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
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:
- $\ds \map f v = \sum_{i \mathop = 1}^n \alpha_i \map {f_i} v$
$\Box$
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 $\alpha_i = 0$.
$\blacksquare$