Basis of a Vector Space

Introduction

A basis is one of the most important ideas in linear algebra. It gives us a way to describe an entire vector space using only a small, carefully chosen set of vectors. If you already understand linear independence, you’re in a perfect position to learn what a basis is and why it matters.

This section will guide you through:

What Is a Basis?

A basis of a vector space is a set of vectors that satisfies two key properties:

Put differently:

Some helpful points:

Examples of Bases

1. Basis of $\mathbb{R}^2$

Common example:

These vectors:

Any vector $(a,b)$ can be written as: $$(a,b) = a(1,0) + b(0,1)$$

2. Another Basis of $\mathbb{R}^2$

The vectors $(1,1)$ and $(1,-1)$ also form a basis.

Why?

3. Basis of $\mathbb{R}^3$

A standard basis:

But many other sets of three independent vectors also work.

How to Check Whether a Set Is a Basis

To check whether a set of vectors forms a basis:

1. Check independence

2. Check spanning

3. Count the vectors

In $\mathbb{R}^n$:

This counting rule is extremely powerful.

Why Bases Matter

A basis allows us to:

A basis is the “skeleton” of a vector space.

Calculator

Testing for linear independence

  • Checks wether whether a set of vectors are independent.
areLinearlyIndependent([[1,0,0], [0,1,0], [0,0,1]]) areLinearlyIndependent([[1,2], [2,4]])

Exercises

Exercises

  1. Determine whether the set $\{(1,2),(3,6)\}$ is a basis for $\mathbb{R}^2$.

    Solution

    The vectors $(1,2)$ and $(3,6)$ are dependent because $(3,6)=3(1,2)$.
    A dependent set cannot be a basis.
    Answer: Not a basis.

  2. Does the set $\{(1,0,1),(0,1,1),(1,1,2)\}$ form a basis for $\mathbb{R}^3$?

    Solution

    Check independence:
    Compute determinant of the matrix with these vectors as rows or columns: $$\begin{vmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 2 \end{vmatrix} = 1(1\cdot 2 - 1\cdot 1) - 0 + 1(0\cdot 1 - 1\cdot 1) = 1(1) + 1(-1) = 0$$ Determinant is $0$ → dependent → not a basis.

  3. Write the vector $(4,5)$ as a linear combination of the basis $\{(1,1),(1,-1)\}$.

    Solution

    Solve: $$(4,5) = a(1,1) + b(1,-1)$$ This gives:

    • $a + b = 4$
    • $a - b = 5$

    Add equations: $2a = 9 \Rightarrow a = 4.5$
    Then $b = 4 - a = -0.5$

    Answer: $(4,5)=4.5(1,1)-0.5(1,-1)$.

  4. True or false: Any set of three independent vectors in $\mathbb{R}^3$ forms a basis.

    Solution

    True.
    In $\mathbb{R}^3$, any set of three independent vectors automatically spans the space.

  5. Determine whether the vectors $(1,0,0)$, $(1,1,0)$, and $(1,1,1)$ form a basis for $\mathbb{R}^3$.

    Solution

    Form matrix and compute determinant: $$\begin{vmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{vmatrix} = 1(1\cdot 1 - 0\cdot 1) = 1$$ Determinant is nonzero → independent → basis.

  6. Express $(2,3,4)$ in terms of the standard basis of $\mathbb{R}^3$.

    Solution

    In the standard basis: $$(2,3,4)=2(1,0,0)+3(0,1,0)+4(0,0,1)$$

  7. Determine whether the set $\{(2,1),(1,2)\}$ spans $\mathbb{R}^2$.

    Solution

    Compute determinant: $$\begin{vmatrix} 2 & 1 \\ 1 & 2 \end{vmatrix} = 4 - 1 = 3$$ Nonzero → independent → spans $\mathbb{R}^2$ → basis.