Condition for Linear Dependence of Linear Functionals in terms of Kernel
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:
- $f \in \span \set {g_1, \ldots, g_n}$
$\blacksquare$
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- 1990: John B. Conway: A Course in Functional Analysis (2nd ed.) ... (previous) ... (next): Appendix $\text{A}$ Preliminaries: $\S 1.$ Linear Algebra: Proposition $1.4$