Mathwords logoMathwords

Vector Basis — Definition, Formula & Examples

A vector basis is a set of linearly independent vectors that spans an entire vector space, meaning every vector in that space can be written as a unique linear combination of the basis vectors.

A set B={v1,v2,…,vn}\mathcal{B} = \{\mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n\} is a basis for a vector space VV if (1) the vectors in B\mathcal{B} are linearly independent and (2) every vector v∈V\mathbf{v} \in V can be expressed as v=c1v1+c2v2+⋯+cnvn\mathbf{v} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_n\mathbf{v}_n for unique scalars c1,c2,…,cnc_1, c_2, \dots, c_n. The number nn is the dimension of VV.

Key Formula

v=c1v1+c2v2+⋯+cnvn\mathbf{v} = c_1\mathbf{v}_1 + c_2\mathbf{v}_2 + \cdots + c_n\mathbf{v}_n
Where:
  • v\mathbf{v} = Any vector in the vector space
  • v1,…,vn\mathbf{v}_1, \dots, \mathbf{v}_n = The basis vectors
  • c1,…,cnc_1, \dots, c_n = Unique scalar coefficients (coordinates relative to the basis)

How It Works

To verify that a set of vectors forms a basis, you check two conditions: linear independence and spanning. For Rn\mathbb{R}^n, a set of nn vectors forms a basis if and only if the matrix formed by placing them as columns has a nonzero determinant. The standard basis for R3\mathbb{R}^3, for instance, is {e1,e2,e3}\{\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3\} where e1=(1,0,0)\mathbf{e}_1 = (1,0,0), e2=(0,1,0)\mathbf{e}_2 = (0,1,0), and e3=(0,0,1)\mathbf{e}_3 = (0,0,1). Any other valid basis for R3\mathbb{R}^3 must also contain exactly three linearly independent vectors.

Worked Example

Problem: Show that the set {(1,2),(3,1)}\{(1, 2), (3, 1)\} is a basis for R2\mathbb{R}^2, then express the vector (7,8)(7, 8) in terms of this basis.
Check linear independence: Form a matrix with these vectors as columns and compute its determinant.
det⁡(1321)=(1)(1)−(3)(2)=−5≠0\det\begin{pmatrix} 1 & 3 \\ 2 & 1 \end{pmatrix} = (1)(1) - (3)(2) = -5 \neq 0
Confirm basis: Since the determinant is nonzero, the two vectors are linearly independent. Two linearly independent vectors in R2\mathbb{R}^2 always span R2\mathbb{R}^2, so this is a basis.
Find coordinates of (7, 8): Solve c1(1,2)+c2(3,1)=(7,8)c_1(1,2) + c_2(3,1) = (7,8), which gives the system c1+3c2=7c_1 + 3c_2 = 7 and 2c1+c2=82c_1 + c_2 = 8. From the second equation, c2=8−2c1c_2 = 8 - 2c_1. Substituting: c1+3(8−2c1)=7c_1 + 3(8 - 2c_1) = 7, so −5c1=−17-5c_1 = -17, giving c1=175c_1 = \frac{17}{5} and c2=65c_2 = \frac{6}{5}.
(7,8)=175(1,2)+65(3,1)\left(7, 8\right) = \frac{17}{5}(1,2) + \frac{6}{5}(3,1)
Answer: The set is a basis for R2\mathbb{R}^2, and (7,8)=175v1+65v2(7, 8) = \frac{17}{5}\mathbf{v}_1 + \frac{6}{5}\mathbf{v}_2.

Why It Matters

Bases are central to linear algebra courses and appear whenever you need coordinate representations, change-of-basis transformations, or dimensional analysis. In computer graphics and machine learning, choosing the right basis (such as eigenvectors for PCA) directly determines how efficiently data is represented and processed.

Common Mistakes

Mistake: Assuming any spanning set is a basis without checking linear independence.
Correction: A spanning set can contain redundant vectors. You must verify that no vector in the set can be written as a linear combination of the others. For Rn\mathbb{R}^n, a basis requires exactly nn linearly independent vectors.