AcademySubspaces And Dimension
Academy
Spanning Sets
Level 1 - Linear Algebra topic page in Subspaces And Dimension.
Principle
A spanning set is a collection of vectors whose linear combinations produce every vector in a target set. To say that vectors span a subspace means they are enough building blocks to reach every vector in that subspace.
Spanning is about coverage.
Notation
\(\operatorname{span}(\mathbf v_1,\ldots,\mathbf v_k)\)
the set of all linear combinations of the listed vectors
\(\mathbf v_1,\ldots,\mathbf v_k\)
vectors used as spanning vectors
\(c_1,\ldots,c_k\)
real coefficients in a linear combination
\(W\)
a subspace being spanned
\(\mathbf b\)
a target vector being tested for membership in a span
The Core Method
The span of vectors \(\mathbf v_1,\ldots,\mathbf v_k\) is
Span definition
\[\operatorname{span}(\mathbf v_1,\ldots,\mathbf v_k)=\{c_1\mathbf v_1+\cdots+c_k\mathbf v_k:c_1,
\ldots,c_k\in\mathbb R\}\]
To test whether \(\mathbf b\) is in the span, solve
Membership equation
\[c_1\mathbf v_1+\cdots+c_k\mathbf v_k=\mathbf b\]
Equivalently, use the matrix with the spanning vectors as columns:
Column span test
\[A\mathbf c=\mathbf b\]
where \(A=(\mathbf v_1\;\cdots\;\mathbf v_k)\). The vector \(\mathbf b\) lies in the span exactly when this system is consistent.
Worked Cases
Question
Decide whether \(\mathbf b=(5,1)\) lies in \(\operatorname{span}((1,2),(3,-1))\).
Answer
Solve \(c_1(1,2)+c_2(3,-1)=(5,1)\). This gives equations \(c_1+3c_2=5\) and \(2c_1-c_2=1\). From the first equation, \(c_1=5-3c_2\). Substitute into the second: \(2(5-3c_2)-c_2=1\), so \(10-7c_2=1\). Then \(-7c_2=-9\), giving \(c_2=9/7\). Hence \(c_1=5-27/7=8/7\). A solution exists, so \(\mathbf b\) lies in the span.
Examples
Question
What is \(\operatorname{span}((4,-2))\)?
Answer
The span is all scalar multiples of \((4,-2)\): \(\{t(4,-2):t\in\mathbb R\}\). Geometrically, this is the line through the origin in the direction \((4,-2)\).