Linear algebra becomes powerful when finite lists of vectors are treated as systems of generators and relations. Span records what can be built; linear independence records whether the building instructions are unique.
Linear combinations and span
Definition. Given vectors \(v_1,\ldots,v_n\in V\), a linear combination is a vector \[ a_1v_1+\cdots+a_nv_n, \] with coefficients \(a_i\in F\). The span of the list is the set of all such linear combinations. A list spans \(V\) if every vector in \(V\) is obtained this way.
Example. In \(\mathbb R^3\), the vectors \(e_1,e_2\) span the plane \(\{(x,y,0):x,y\in\mathbb R\}\), not all of \(\mathbb R^3\). In \(F_2[x]\), the polynomials \(1,x,x^2\) span every polynomial of degree at most \(2\). In \(M_{2,2}(F)\), the four matrix units \(E_{11},E_{12},E_{21},E_{22}\) span the whole matrix space.
Relations among vectors
Definition. A linear relation among \(v_1, \ldots,v_n\) is an equation \[ a_1v_1+\cdots+a_nv_n=0. \] The relation is trivial if all \(a_i=0\). The list is linearly independent if the only relation is the trivial relation; otherwise it is linearly dependent.
Proposition. A list \((v_1,\ldots,v_n)\) is linearly dependent if and only if one of its vectors lies in the span of the others. More precisely, if \(a_j\ne0\) in a nontrivial relation, then \[ v_j=-a_j^{-1}\sum_{i\ne j}a_iv_i. \]
Proof. A nontrivial relation with \(a_j\ne0\) can be solved for \(v_j\), giving membership in the span of the remaining vectors. Conversely, if \(v_j=\sum_{i\ne j}b_iv_i\), then \(v_j-\sum_{i\ne j}b_iv_i=0\) is a nontrivial relation.
Uniqueness of coordinates
Theorem. For a list \((v_1,\ldots,v_n)\), the following are equivalent: the list is linearly independent; every vector in its span has at most one expression as a linear combination of the list; the linear map \(F^n\to V\), \(a\mapsto\sum_i a_iv_i\), is injective.
Proof. If two coefficient vectors give the same linear combination, subtracting the two expressions gives a linear relation. Thus nonunique expression is exactly the same thing as a nontrivial relation. The map formulation says the same fact as \(\ker T=\{0\}\).
The exchange step
Lemma. Suppose \((u_1,\ldots,u_m)\) is linearly independent and lies in \(\operatorname{span}(v_1, \ldots,v_n)\). Then \(m\le n\). More generally, one may replace some of the \(v_i\)'s by the \(u_j\)'s while preserving the span.
Proof. Write \(u_1\) as a linear combination of the spanning list. At least one coefficient is nonzero, so solve for one \(v_i\) and replace it by \(u_1\) without changing the span. Repeat. If the process required more replacements than there are \(v_i\)'s, the \(u_j\)'s would be dependent, contradicting independence.
Computing span and independence
Calculation. For vectors in \(F^m\), place the vectors as columns of a matrix \(A\). The span is the column space \(\operatorname{im}A\). The list is independent exactly when \(Ax=0\) has only the zero solution, i.e. exactly when every column is a pivot column after row reduction.
Counterexample. Three nonzero vectors in \(\mathbb R^3\) need not span \(\mathbb R^3\). For instance \((1,0,0),(0,1,0),(1,1,0)\) are all contained in the plane \(z=0\) and are linearly dependent. Counting vectors is not enough; one must test relations.