# Span and Linear Combinations

Mathematics I · Linear Algebra · https://tryals.app/learn/mathematics-i/span-and-linear-combinations

## Building with Vectors

A **linear combination** of vectors is an expression built by scaling and adding: $c_1 v_1 + c_2 v_2 + \dots + c_k v_k$, where $c_i$ are scalars. It is the fundamental way to construct new vectors.

The **span** $\operatorname{span}\{v_1, \dots, v_k\}$ is the collection of **all** possible linear combinations of those vectors. The span is always a **subspace** because it contains $\mathbf{0}$ and is closed under vector addition and scalar multiplication.

| Vector Set | Geometric Span |
| --- | --- |
| Single nonzero vector | A **line** through the origin |
| Two non-parallel vectors | The **whole plane** |
| Two parallel vectors | Just a **line** (redundant) |

## Spanning and Systems

A **spanning set** for a space $V$ is a set of vectors whose span is all of $V$. For example, $(1, 0)$ and $(0, 1)$ span $\mathbb{R}^2$ since any vector $(a, b) = a(1,0) + b(0,1)$.

To determine if $w$ is in the span of other vectors, solve the **linear system** $w = c_1 v_1 + \dots + c_k v_k$ for unknowns $c_i$. A solution exists if and only if $w$ lies in the span.

> **Common pitfall:** Assuming more vectors always **span more**. Adding a vector that is already a linear combination of the existing ones does **not** enlarge the span. Only a vector pointing in a genuinely new direction expands the reach.

## Practice questions

6 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Adding a vector to a list always enlarges its span.

**Answer:** False

**Why:** False. If the new vector already lies in the span — a multiple of one you have, or any combination of them — the span is unchanged. The span grows only when the vector is *independent* of the existing ones, which is exactly what makes independence the right notion for measuring dimension.

Page: https://tryals.app/practice/mathematics-i/span-and-linear-combinations/adding-a-vector-to-a-list-always-enlarges-its-span

### 2. The span of two vectors in ordinary space is usually a plane, but sometimes only a line. Why does the number of vectors fail to determine the dimension of what they sweep out?

A. Because a plane requires exactly three vectors, so two vectors can never span more than a line.
B. Because the dimension of a span depends on the lengths of the vectors involved.
C. Because span counts independent directions, not vectors.
D. Because spans of two vectors are lines, and only spans containing the zero vector are planes.

**Answer:** C. Because span counts independent directions, not vectors.

**Why:** Counting vectors counts the list; dimension counts the independent directions in it. Every combination of (1,2,1) and (2,4,2) lies on the line the first already swept, the second was free and useless. Which is why dimension is defined via a basis, spanning *and* independent, rather than by counting whatever you happened to write down.

Page: https://tryals.app/practice/mathematics-i/span-and-linear-combinations/the-span-of-two-vectors-in-ordinary-space-is-usually-a-plane-but

### 3. Let W be the span of (1, 0, 1) and (0, 1, 1). Decide for each vector whether it lies in W.

**Answer:**

- In the span: (2, 3, 5), (0, 0, 0), (-1, 4, 3), (7, -7, 0)
- Not in the span: (1, 1, 3), (5, 5, 5)

**Why:** The span consists of the vectors (a, b, a + b), so membership is the single test 'third coordinate equals first plus second'. Checking: 2 + 3 = 5 yes, 1 + 1 = 2 not 3, -1 + 4 = 3 yes, 5 + 5 = 10 not 5, 7 - 7 = 0 yes. The zero vector always qualifies.

Page: https://tryals.app/practice/mathematics-i/span-and-linear-combinations/let-w-be-the-span-of-1-0-1-and-0-1-1-decide-for-each-vector

### 4. The span of a set is always a subspace, whatever vectors you start from. Why is that automatic?

A. Because every set of vectors is itself a subspace
B. Because a span is by construction closed under addition and scaling
C. Because spans are always the whole space
D. Because the span is defined as the smallest subspace, so it is one by definition alone

**Answer:** B. Because a span is by construction closed under addition and scaling

**Why:** The three conditions fall out of the definition. This is why "span" is the standard way to BUILD a subspace: you cannot fail to get one.

Page: https://tryals.app/practice/mathematics-i/span-and-linear-combinations/the-span-of-a-set-is-always-a-subspace-whatever-vectors-you-start

### 5. A linear system $w = c_1 v_1 + \dots + c_k v_k$ has no solution for the scalars $c_i$. What does this algebraic failure reveal geometrically about $w$ relative to the set $\{v_1, \dots, v_k\}$?

A. The set of vectors $\{v_1, \dots, v_k\}$ spans only the origin point
B. The vector $w$ must be orthogonal to every single vector in the set
C. The vector $w$ lies entirely outside the subspace spanned by the set
D. The vector $w$ is parallel to a redundant linear combination in the set

**Answer:** C. The vector $w$ lies entirely outside the subspace spanned by the set

**Why:** Insolubility means no weighting of the vectors can reach $w$, placing it outside their span. It need not be orthogonal or perpendicular to the subspace, merely non-coplanar with the reach of the set.

Page: https://tryals.app/practice/mathematics-i/span-and-linear-combinations/a-linear-system-w-c1-v1-ck-vk-has-no-solution-for-the-scalars

### 6. Order the steps that decide whether a vector w lies in the span of v1 and v2.

**Answer:**

1. Write w = c1 v1 + c2 v2 with the scalars c1 and c2 unknown
2. Turn the vector equation into a linear system, one equation per coordinate
3. Row reduce the system
4. If the system is consistent, w is in the span, and the solution gives the coefficients

**Why:** Every span question becomes a linear system in the coefficients, so the whole Gaussian-elimination machinery applies. Consistency means membership; inconsistency means w points somewhere the given vectors cannot reach.

Page: https://tryals.app/practice/mathematics-i/span-and-linear-combinations/order-the-steps-that-decide-whether-a-vector-w-lies-in-the-span-of-v1