Condition for Linear Dependence of Linear Functionals in terms of Kernel

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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 1

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 $\alpha_i = 0$.

$\blacksquare$


Proof 2

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:

\(\ds \map T {\alpha x + \beta y}\) \(=\) \(\ds \tuple {\map {g_1} {\alpha x + \beta y}, \map {g_2} {\alpha x + \beta y}, \ldots, \map {g_n} {\alpha x + \beta y} }\)
\(\ds \) \(=\) \(\ds \tuple {\alpha \map {g_1} x + \beta \map {g_1} y, \alpha \map {g_2} x + \beta \map {g_2} y, \ldots, \alpha \map {g_n} x + \beta \map {g_n} y}\) $g_i$ is linear
\(\ds \) \(=\) \(\ds \alpha \tuple {\map {g_1} x, \ldots, \map {g_n} x} + \beta \tuple {\map {g_1} y, \ldots, \map {g_n} y}\)
\(\ds \) \(=\) \(\ds \alpha T x + \beta T y\)

so $T$ is linear.

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

That is, if and only if:

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

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 $x \in X$:

\(\ds \map f x\) \(=\) \(\ds \map {\widetilde f} {\map T x}\)
\(\ds \) \(=\) \(\ds \map h {\map T x}\) since $h$ and $\widetilde f$ coincide on $\Img T$
\(\ds \) \(=\) \(\ds \map h {\sum_{i \mathop = 1}^n \map {g_i} x e_i}\)
\(\ds \) \(=\) \(\ds \sum_{i \mathop = 1}^n \map {g_i} x \map h {e_i}\)

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

So we have:

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

$\blacksquare$


Sources


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