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Tuesday, June 9, 2026

MA25C02| | Linear Algebra |Vector Space| Linear Span| Solved example: Do u=(1,1,2), v=(1,0,1), w=(2,1,3) span

Linear Span — Linear Algebra

Linear Span

Understanding which vectors are reachable — and why it matters

Tutorial Video:Linear Span
Linear Combinations Linear Span Linear Algebra Vector Space
Definition

The linear span (or simply span) of a set of vectors \(\{v_1, v_2, \dots, v_k\}\) is the set of all possible linear combinations of those vectors.

$$\operatorname{Span}\{v_1,\dots,v_k\} = \bigl\{\, c_1 v_1 + c_2 v_2 + \cdots + c_k v_k \;\bigm|\; c_1,\dots,c_k \in \mathbb{R} \,\bigr\}$$
Notes
  • \(L(S) \subset V\)
  • If \(S = \emptyset\), then \(L(S) = \{0\}\)

Why is span important?

🧭
Reachability
Determines which vectors are reachable from the origin using given directions.
📐
Basis
A spanning set that is linearly independent forms a basis of the space.
🔢
Linear systems
In \(Ax=b\), the columns of \(A\) span the column space — telling which \(b\) yield a solution.
Worked Example 1
Do \(u=(1,1,2)\), \(v=(1,0,1)\), and \(w=(2,1,3)\) span \(\mathbb{R}^3\)?
1
Criterion. Three vectors span \(\mathbb{R}^3\) if and only if they are linearly independent — equivalently, the determinant of the matrix they form is nonzero.
2
Form the matrix.
\( A = \begin{pmatrix} 1 & 1 & 2 \\ 1 & 0 & 1 \\ 2 & 1 & 3 \end{pmatrix} \)
3
Compute the determinant.
\( \begin{aligned} \det(A) &= 1(0\cdot3 - 1\cdot1) - 1(1\cdot3 - 1\cdot2) + 2(1\cdot1 - 0\cdot2)\\ &= 1(0-1) - 1(3-2) + 2(1-0)\\ &= -1 - 1 + 2 \\[4pt] &= 0 \end{aligned}\)
4
Interpret. Since \(\det(A) = 0\), the vectors are linearly dependent. In fact, \(w = u + v\), so \(w\) adds no new direction.
\(u,\, v,\, w\) do not span \(\mathbb{R}^3\).
The three vectors lie in a 2-dimensional plane (a subspace of \(\mathbb{R}^3\)), not all of \(\mathbb{R}^3\). Any vector outside that plane cannot be reached as a linear combination of \(u, v, w\).

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