← mapVector Spaces

Span & Linear Combinations

⚗ Dr. Möbius, from the lab

I'm going to tell you a secret that linear algebra spends a whole semester pretending is many ideas: there is only one move. Scale some vectors, add the results. That's it. That's the entire goddamn art form. Everything — span, basis, rank, eigenvectors, the works — is just that one move asked in different accents. Today we name it, and then we name everything it can reach.

THE BIG IDEA

A linear combination scales vectors and adds them; the span of a set is all such combinations, and it is always the smallest subspace containing the set.

The only move in the building

Take vectors $v_1, v_2, \dots, v_k$ in a vector space $V$ and scalars $c_1, c_2, \dots, c_k \in \mathbb{R}$ . The expression $c_1 v_1 + c_2 v_2 + \cdots + c_k v_k$ is a linear combination of the $v_i$ . Scale each vector, add them all up. The $c_i$ are the coefficients or weights.

I am not exaggerating when I say this is the only operation in linear algebra. Matrix multiplication is built from it. Solving systems is reverse-engineering it. The whole Subspaces node was secretly about which sets are closed under it (because closure under $+$ and scaling together is exactly closure under linear combinations). Burn the phrase into your skull: linear algebra is the study of linear combinations and nothing else. If you think something else is happening, look again — it's a linear combination wearing a hat.

Span: everything you can reach

Given a set $S = \{v_1, \dots, v_k\}$ , the span of $S$ is the set of all linear combinations you can form from it: $\operatorname{span}(S) = \{ c_1 v_1 + \cdots + c_k v_k : c_1, \dots, c_k \in \mathbb{R} \}.$ It's every destination reachable by scaling and adding. If you hand me three vectors, $\operatorname{span}$ is the complete map of where those three can take you.

Span is a subspace (and here's the proof)

This is the theorem that makes span worth defining. Reach for the three-condition test from the Subspaces node.

Theorem. For any $v_1, \dots, v_k \in V$ , $\operatorname{span}(v_1, \dots, v_k)$ is a subspace of $V$ .

Proof. Let $W = \operatorname{span}(v_1, \dots, v_k)$ .

Contains $\mathbf{0}$ . Take all coefficients zero: $0v_1 + \cdots + 0v_k = \mathbf{0}$ (using $0v = \mathbf{0}$ from the Vector Spaces node). So $\mathbf{0} \in W$ . ✓
Closed under addition. Take two elements of $W$ : $\mathbf{x} = a_1 v_1 + \cdots + a_k v_k, \qquad \mathbf{y} = b_1 v_1 + \cdots + b_k v_k.$ Add them and group by like vectors (commutativity + the distributive axioms): $\mathbf{x} + \mathbf{y} = (a_1 + b_1)v_1 + \cdots + (a_k + b_k)v_k.$ That's again a linear combination of the $v_i$ , so $\mathbf{x} + \mathbf{y} \in W$ . ✓
Closed under scaling. For $c \in \mathbb{R}$ : $c\mathbf{x} = c(a_1 v_1 + \cdots + a_k v_k) = (ca_1)v_1 + \cdots + (ca_k)v_k \in W. \;✓$

All three conditions hold, so $W$ is a subspace. $\blacksquare$

In fact $\operatorname{span}(S)$ is the smallest subspace containing $S$ : any subspace that contains all the $v_i$ must (being closed under combinations) contain every linear combination of them, hence all of $\operatorname{span}(S)$ . Span is the tightest possible subspace wrapper around your vectors.

The geometry: lines, planes, and "is $w$ in there?"

Picture it in $\mathbb{R}^3$ :

Span of one nonzero vector $v$ : all scalar multiples $cv$ — a line through the origin.
Span of two non-parallel vectors $u, v$ : a plane through the origin.
Span of two parallel vectors: collapses back to a line (the second adds nothing — hold that thought, it's the whole next node).

Drag the two vectors below and the slider weights; the dot is the linear combination $c_1 u + c_2 v$ . Watch which points it can and can't reach — that reachable set is the span.

vector playground — linear combinations

c₁ = 1c₂ = 1

u = (2, 1)v = (-1, 2)c₁u + c₂v = (1, 3)

Now the question that powers the rest of the course: is a given vector $w$ in $\operatorname{span}(v_1, \dots, v_k)$ ? By definition, that asks: do there exist coefficients $c_1, \dots, c_k$ with $c_1 v_1 + \cdots + c_k v_k = w?$ That is a system of linear equations in the unknowns $c_i$ . And solving systems is exactly what Gaussian elimination from the Matrices stratum does. So: $w \in \operatorname{span}(v_1, \dots, v_k) \iff \text{the system } c_1 v_1 + \cdots + c_k v_k = w \text{ has a solution.}$ The abstract membership question cashes out as elimination. This is the marriage again: a Sets-flavored "is it in the set?" answered by a Matrices-flavored row reduction.

Column space: the matrix's own span

Here's where it gets gorgeous. Stack vectors $v_1, \dots, v_n$ as the columns of a matrix $A$ . Then the product $A\mathbf{x}$ is, by the definition of matrix–vector multiplication (Matrices stratum), exactly $A\mathbf{x} = x_1 v_1 + x_2 v_2 + \cdots + x_n v_n$ — a linear combination of the columns, weighted by the entries of $\mathbf{x}$ . So the set of all possible outputs $A\mathbf{x}$ is the span of the columns. We name it the column space: $\operatorname{Col}(A) = \operatorname{span}(\text{columns of } A).$

And the headline theorem of solvability — and I want you to feel the weight of this: $A\mathbf{x} = \mathbf{b} \text{ has a solution} \iff \mathbf{b} \in \operatorname{Col}(A).$ Read that in plain English: $A\mathbf{x} = \mathbf{b}$ is solvable exactly when $\mathbf{b}$ can be mixed from the columns of $A$ . "Solving a system" means "expressing $\mathbf{b}$ as a mixture of columns." You've been computing spans since the Matrices stratum without knowing the word — and that should feel a little bit like finding out your name in a foreign language.

Spanning sets, and a teaser

A set $S$ spans $V$ if $\operatorname{span}(S) = V$ — every vector in $V$ is reachable. For $\mathbb{R}^n$ , the standard basis $e_1, \dots, e_n$ (the columns of the identity matrix) spans, since $(x_1, \dots, x_n) = x_1 e_1 + \cdots + x_n e_n$ . You need at least $n$ vectors to span $\mathbb{R}^n$ .

But here's the rot we'll exterminate next node: some vectors in a spanning set are redundant — they're already in the span of the others, so deleting them changes nothing. Two parallel vectors only span a line; the second is pure deadweight. Detecting and removing that deadweight is the entire job of Linear Independence. File it away — we're going hunting.

🔬 SPECIMENS (worked examples)

Worked example 1 — is this vector in the span?▸

Is $w = (4, 5)$ in $\operatorname{span}\big((1, 1), (2, 3)\big)$ ? If so, give the coefficients.

We need $c_1, c_2$ with $c_1(1,1) + c_2(2,3) = (4,5)$ . Componentwise that's the system $\begin{cases} c_1 + 2c_2 = 4 \\ c_1 + 3c_2 = 5 \end{cases}$ Subtract the first equation from the second: $c_2 = 1$ . Back-substitute: $c_1 + 2(1) = 4 \Rightarrow c_1 = 2$ .

Check: $2(1,1) + 1(2,3) = (2,2) + (2,3) = (4,5)$ . ✓

So yes, $w \in \operatorname{span}\big((1,1),(2,3)\big)$ with coefficients $c_1 = 2, c_2 = 1$ . The membership question became a $2\times 2$ system and Gaussian elimination closed it — exactly the marriage of strata I keep harping on.

Worked example 2 — column space and the brutal truth about solvability▸

Let $A = \begin{pmatrix} 1 & 2 \\ 2 & 4 \end{pmatrix}$ . Is $\mathbf{b} = (3, 7)$ in $\operatorname{Col}(A)$ ? What about $\mathbf{b}' = (3, 6)$ ?

$\operatorname{Col}(A) = \operatorname{span}\big((1,2), (2,4)\big)$ . But $(2,4) = 2(1,2)$ — the second column is parallel to the first! So the column space is just the line $\operatorname{span}\big((1,2)\big) = \{(t, 2t) : t \in \mathbb{R}\}$ .

A vector $(x, y)$ is in this line exactly when $y = 2x$ .

For $\mathbf{b} = (3, 7)$ : is $7 = 2(3) = 6$ ? No. So $\mathbf{b} \notin \operatorname{Col}(A)$ , and $A\mathbf{x} = (3,7)$ has no solution.

For $\mathbf{b}' = (3, 6)$ : is $6 = 2(3)$ ? Yes. So $\mathbf{b}' \in \operatorname{Col}(A)$ , and $A\mathbf{x} = (3,6)$ is solvable (e.g. $\mathbf{x} = (3, 0)$ , giving $3(1,2) = (3,6)$ ). The redundant second column shrank the reachable set to a line — a preview of why redundancy matters.

Worked example 3 — the parallel-vector trap: two vectors that span exactly nothing new▸

Do $(1, 2)$ and $(2, 4)$ span $\mathbb{R}^2$ ? Most students say yes because "two vectors in $\mathbb{R}^2$ ." Settle it.

The trap is counting vectors instead of checking reach. $(2,4) = 2 \cdot (1,2)$ , so the two are parallel — the second adds nothing new. Their span is $\operatorname{span}\big((1,2),(2,4)\big) = \operatorname{span}\big((1,2)\big) = \{(t, 2t)\},$ which is a single line through the origin, not the whole plane.

Concretely, $(1, 0)$ is not reachable: $c_1(1,2) + c_2(2,4) = (c_1 + 2c_2)(1, 2)$ always lies on the line $y = 2x$ , and $(1,0)$ has $0 \ne 2$ . So they do not span $\mathbb{R}^2$ . To span $\mathbb{R}^2$ you need two vectors pointing in genuinely different directions. Number of vectors is necessary, not sufficient — direction is what counts, which is precisely the independence question coming next.

☠ KNOWN HAZARDS

Confusing "span" the verb with "span" the noun. A set of vectors spans (verb) a space; $\operatorname{span}(S)$ (noun) is the subspace they reach. " $S$ spans $\mathbb{R}^3$ " means $\operatorname{span}(S) = \mathbb{R}^3$ . This is pure vocabulary and getting it wrong reads as sloppy.
Thinking more vectors always means a bigger span. Adding a vector already in the span changes nothing — it's deadweight, and we will hunt it in the next node. Two parallel vectors span the same line one does. Quantity isn't reach.
Forgetting span always contains $\mathbf{0}$ . Set all coefficients to zero. Any "span" that supposedly misses the origin is a contradiction — you mis-set up the problem.
Reading $A\mathbf{x}=\mathbf{b}$ as anything but "mix the columns to hit $\mathbf{b}$ ". That column-mixture picture is the whole meaning of solvability; lose it and the column space stays mysterious forever.

TL;DR

▸
A linear combination is $c_1 v_1 + \cdots + c_k v_k$ — scale and add. It's the only operation linear algebra ever performs.
▸
$\operatorname{span}(S)$ is the set of all linear combinations of $S$ , and it is always a subspace — in fact the smallest subspace containing $S$ (proved via the three-condition test).
▸
Geometrically: one vector spans a line through $\mathbf{0}$ , two non-parallel vectors span a plane. "Is $w \in \operatorname{span}(\dots)$ ?" $\iff$ "does a linear system have a solution?" — Gaussian elimination answers it.
▸
The column space $\operatorname{Col}(A) = \operatorname{span}(\text{columns})$ , and $A\mathbf{x}=\mathbf{b}$ is solvable $\iff \mathbf{b} \in \operatorname{Col}(A)$ — i.e. $\mathbf{b}$ is a mixture of the columns.
▸
A set spans $V$ when its span is all of $V$ . Spanning sets can carry redundant vectors — the cleanup problem of the next node.

unlocks

Linear Independence →