span + linear dependence

#linear-algebra #span #subspace #vector-space

def (Linear Combination)

A linear combination of $v_{1} \dots v_{m} \in V$ is a vector of the form

a_{1} v_{1} + \dots + v_{m} v_{m} $ $ w h e r e $ a_{1} \dots a_{m} \in F $

def (Span)

The set of all possible linear combinations of $v_{1} \dots v_{m} \in V$ is called the span of vectors $v_{1} \dots v_{m}$ . Denoted $span (v_{1} \dots v_{m})$

{a_{1} v_{1} + \dots + a_{m} v_{m} | a_{1} \dots a_{m} F}

By convention $span (ϕ) = {0}$

Proposition: The set

span (v_{1} \dots v_{m})

is the smallest subspace of

V

containing

v_{1} \dots v_{m}

Proof: let $S = span (v_{1} \dots v_{m})$

S in a subspace:

$0 \in S$ because $a_{1} = \dots = a_{m} = 0 (\in F)$ is in the span
$S$ is closed under addition. Suppose $u, w \in S$ , then $\exists a_{1} \dots a_{m}$ and $b_{1} \dots b_{m} \in F$ .
$\begin{aligned} u & = a_{1} v_{1} + \dots + a_{m} v_{m} \\ w & = b_{1} v_{1} + \dots + b_{m} v_{m} \\ u + w & = (a_{1} + b_{1}) v_{1} + \dots + (a_{m} + b_{m}) v_{m} \in S \end{aligned}$
$S$ is closed under scalar multiplication. Suppose $u \in S$ . Then,
$\begin{aligned} u & = a_{1} v_{1} + \dots + a_{m} v_{m} \\ α u = (α a_{1}) v_{1} + \dots + (α a_{m}) v_{m} \end{aligned}$
Thus, $S = span (v_{1} \dots v_{m})$ is a subspace.

$S$ is the smallest subspace

Outline: Show $v_{1} \dots v_{m} \in S$ and if $T$ is any subspace s.t. $v_{1} \lodts v_{m} \in T$ then $s \subseteq T$ .
- For each $v_{i}$ , $v_{i} = 0 v_{1} + \dots + 1 v_{i} + \dots + 0 v_{m}$ .
- Suppose $T$ is a subspace containing $v_{1} \dots v_{m}$ . Since $T$ is a subspace it is closed under addition and scalar multiplication i.e. it is closed under linear combination. Thus $T$ contains all linear combinations of $v_{1} \dots v_{m} ⟹ S \subseteq T$ .

If $span (v_{1} \dots v_{m}) = V$ then we say $v_{1} \dots v_{m}$ spans $V$ .
A vector space is said to be finite dimensional if it is spanned by finitely many vectors. Otherwise, we say it is infinite-dimensional.

def (Linearly Independent)

A subspace of $V$ is linearly independent iff

a_{1} v_{1} + \dots + a_{n} v_{n} = 0 ⟹ a_{1} = \dots = a_{n} = 0

By convention, $ϕ$ is linearly independent.

Proposition: A subspace is linearly independent iff none of the vectors is a linear combination of the other.

Proof: In both cases we prove the contrapositive

[=>] W.L.O.G. let $v_{1}$ be a linear combination of $v_{2} \dots v_{m}$ . Thus,

\begin{aligned} v_{1} & = a_{2} v_{2} + \dots a_{n} v_{n} \\ 0 = (- 1) v_{1} + a_{2} v_{2} + \dots a_{n} v_{n} \end{aligned}

Thus, there is a non-trivial way to represent 0.

[<=] Suppose $v_{1} \dots v_{n}$ are linearly dependent. Then there exists $a_{1} \dots a_{n}$ not all zero s.t. $0 = a_{1} v_{1} + \dots a_{n} v_{n}$ . W.L.O.G. $a_{1} \neq 0$

v_{1} = - \frac{a_{2}}{a_{1}} v_{2} \dots - \frac{a_{n}}{a_{1}}

Thus one of the vectors is a linear combination of the other.

Lemma (Linear Independence Lemma)

Suppose $v_{1} \dots v_{n}$ is a linearly dependent set $V$ . Then $\exists j \in [1, n]$ s.t. the following hold:

$v_{j} \in span (v_{1} \dots v_{j - 1})$
If $v_{j}$ is removed, the remaining span equals the original span