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 Incoming links: Linear span

A family is linearly independent if and only if every vector in its linear span has unique coefficients as a linear combination of that family. For a short proof, subtract two representations to get

\sum_{i} (a_{i} - b_{i}) v_{i} = 0

; linear independence forces each difference to vanish. Conversely, uniqueness applied to the zero vector forces every linear relation to use only zero coefficients. This explains the practical role of linear independence: a coordinate vector relative to an ordered basis is unambiguous.

c

p = proj_{c} a = \frac{a \cdot c}{c \cdot c} c, r = a - p, r \cdot c = 0.

For

a = (3, 4)

and

c = (2, 0)

, this gives

p = (6/4) (2, 0) = (3, 0)

and

r = (0, 4)

r \cdot c = 0

c

(- 2, 0)

gives the same

p = (- 6/4) (- 2, 0) = (3, 0)

: reversing direction changes the scalar projection, but preserves the vector projection onto the line.

For a short proof of the nearest-point property, any vector

z

on this line satisfies

∥ a - z ∥^{2} = ∥ r ∥^{2} + ∥ p - z ∥^{2}

by the Pythagorean identity. The second term is uniquely minimized at

z = p

. This connects a directional measurement to the orthogonal projection used in approximation.^[ref]

V

F

A \subseteq V

extends to a basis. This is an algebraic Hamel basis: each representation uses a finite linear combination, even when the basis is infinite.^[ref]

A

P_{A} = {S \subseteq V : A \subseteq S, S is linearly independent over F}, S \leq T ⟺ S \subseteq T .

It is a set, being a subset of the power set of

V

, and is nonempty because

A \in P_{A}

A short proof outline supplies both essential steps:

For a nonempty chain in a partial order $C \subseteq P_{A}$ , put $U = ⋃_{S \in C} S$ . Any finite linear relation in $U$ lies in one member of $C$ : among finitely many containing members, one contains the rest. Its linear independence makes every coefficient zero. Thus $U \in P_{A}$ is an upper bound. For the empty chain in a partial order, use $A$ ; its empty union need not contain $A$ .
Zorn's lemma gives a maximal element $B$ . If a vector $v$ lies outside its linear span, then $B \cup {v}$ is linearly independent: a linear relation with nonzero coefficient on $v$ would express $v$ in that linear span. This contradicts maximality, so $B$ is a spanning set and hence a basis.^[ref]

Taking

A = \emptyset

proves existence, including

V = {0}

, whose basis is the empty set. The chain in a partial order need not be indexed by the natural numbers: only the finitely many members needed for a particular linear relation are compared. An arbitrary union of linearly independent sets can fail:

{v}

and

{2 v}

for a nonzero vector over the real numbers are each linearly independent, but their union is linearly dependent. The nesting of a chain in a partial order is what prevents this failure.