Linear Combinations and Span

Linear combination

A linear combination of vectors $\mathbf{v}_1, \mathbf{v}_2, \ldots, \mathbf{v}_k$ is any vector of the form $c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_k \mathbf{v}_k$

with scalars $c_i \in \mathbb{R}$ (called coefficients). Zero coefficients are allowed.

Span

The span of $\{\mathbf{v}_1, \ldots, \mathbf{v}_k\}$ is the set of all linear combinations: $\operatorname{Span}\{\mathbf{v}_1, \ldots, \mathbf{v}_k\} = \{ c_1 \mathbf{v}_1 + \cdots + c_k \mathbf{v}_k : c_i \in \mathbb{R}\}$

The span is always a subspace of the ambient vector space (closed under addition and scalar multiplication, and contains $\mathbf{0}$ via the zero coefficients).

In $\mathbb{R}^2$:

• Span of a single non-zero vector: a line through origin.
• Span of two non-parallel vectors: the whole plane $\mathbb{R}^2$.

Step-by-step — express [7, 4] via [1, 2] and [3, −1]

We want $c_1, c_2 \in \mathbb{R}$ with $c_1 \begin{pmatrix} 1 \\ 2 \end{pmatrix} + c_2 \begin{pmatrix} 3 \\ -1 \end{pmatrix} = \begin{pmatrix} 7 \\ 4 \end{pmatrix}$

This is equivalent to the linear system $c_1 + 3 c_2 = 7 \quad \text{(first component)}$ $2 c_1 - c_2 = 4 \quad \text{(second component)}$

Step 1 — Solve. Multiply the first equation by 2: $2 c_1 + 6 c_2 = 14$

Subtract the second equation: $(2 c_1 + 6 c_2) - (2 c_1 - c_2) = 14 - 4$ $7 c_2 = 10 \implies c_2 = \dfrac{10}{7}$

Step 2 — Back substitute. $c_1 = 7 - 3 \cdot \dfrac{10}{7} = \dfrac{49 - 30}{7} = \dfrac{19}{7}$

Solution: $\boxed{\mathbf{v} = \tfrac{19}{7} (1, 2) + \tfrac{10}{7} (3, -1)}$.

Verification

$\tfrac{19}{7}(1, 2) + \tfrac{10}{7}(3, -1) = \left(\tfrac{19 + 30}{7}, \tfrac{38 - 10}{7}\right) = (7, 4)$ ✓

Geometric intuition

Linear combinations are about scaling and adding:

• $c_1 \mathbf{v}_1$: stretch $\mathbf{v}_1$ by factor $c_1$.
• $c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2$: translate the tip of $c_1 \mathbf{v}_1$ by $c_2 \mathbf{v}_2$.

In this example, we're reaching $(7, 4)$ by walking $19/7$ steps in the $(1, 2)$ direction, then $10/7$ steps in the $(3, -1)$ direction.

When span is everything

Two vectors in $\mathbb{R}^2$ span $\mathbb{R}^2$ if and only if they are not parallel. For $(1, 2)$ and $(3, -1)$: are they parallel? Ratio check: $1/3 \ne 2/-1$, so non-parallel. Their span is the entire plane — which is why we could represent any target vector, including $(7, 4)$, as a linear combination.

If instead we tried $\mathbf{v}_1 = (1, 2)$ and $\mathbf{v}_2 = (2, 4)$ (parallel), their span would be just the line $y = 2x$, and only points on that line would be reachable.

Common mistakes

• Assuming the combination is always unique. If the vectors span but are redundant (e.g. three vectors in $\mathbb{R}^2$), the coefficients are not unique — there are infinitely many.
• Forgetting the span must contain the zero vector. It does, using all coefficients equal to zero.
• Treating "linear combination" as just adding vectors. It's scaling then adding — the scalars matter.

Try it in the visualization

Sliders for $c_1, c_2$ build up the linear combination $c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2$ step by step. The tip traces out the span as you sweep through values, revealing a line (one vector) or plane (two non-parallel vectors).