Basis, Coordinates and Dimension of Vector Spaces

Before we formally define the basis of a vector space, we examine examples of bases in two-dimensional space from physics and analytical geometry to build foundational intuition.

Definition of a Basis for 2-Dimensional Space Using Rectangular Axes

We first discuss what we know about vectors in a two-dimensional space, as used in physics and analytical geometry. The graph below shows a point \( P \), the orthogonal unit vectors \( \mathbf{i} \) and \( \mathbf{j} \), and the vector \( \mathbf{OP} \).

Rectangular coordinate system showing unit vectors i and j, point P, and vector OP.

Vector \( \mathbf{OP} \) may be expressed as a linear combination of the unit vectors \( \mathbf{i} \) and \( \mathbf{j} \):

\( \mathbf{OP} = x\,\mathbf{i} + y\,\mathbf{j} \)

Here \( x \) and \( y \) are the coordinates of \( \mathbf{OP} \) relative to the basis \( \{ \mathbf{i}, \mathbf{j} \} \).

Using vector addition via a parallelogram, there is a unique way to combine multiples of \( \mathbf{i} \) and \( \mathbf{j} \) to obtain \( \mathbf{OP} \). Every vector in the plane has a unique representation as \( x\mathbf{i} + y\mathbf{j} \), provided \( \mathbf{i} \) and \( \mathbf{j} \) are not parallel (equivalently, linearly independent). Since every vector in the plane can be so expressed, the pair \( \{ \mathbf{i}, \mathbf{j} \} \) forms a basis for the 2-dimensional space.

Generalizing the Definition of a Basis for a 2-Dimensional Space

We now extend the concept to a basis made of vectors that are not necessarily orthogonal or unit vectors. Let \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) be any two non-parallel vectors in the plane.

Non-rectangular coordinate system showing basis vectors v1 and v2, and vector OP.

Using a parallelogram, there is a unique way to combine multiples of \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) to obtain \( \mathbf{OP} \). That is, there is a unique representation

\( \mathbf{OP} = a\,\mathbf{v}_1 + b\,\mathbf{v}_2 \)

The scalars \( a \) and \( b \) are the coordinates of \( \mathbf{OP} \) relative to the basis \( \{ \mathbf{v}_1, \mathbf{v}_2 \} \), provided \( \mathbf{v}_1 \) and \( \mathbf{v}_2 \) are not parallel (i.e., linearly independent). Thus \( \{ \mathbf{v}_1, \mathbf{v}_2 \} \) is a basis for the two-dimensional space.

Generalizing the Definition of a Basis for a Subspace

Let \( S = \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \} \) be a set of vectors in a vector space \( V \). Then \( S \) is a basis for \( V \) if and only if:

  1. the vectors in \( S \) are linearly independent;
  2. the vectors in \( S \) span \( V \).
Key idea. A basis is a minimal spanning set — no vector can be removed without losing the span — and a maximal linearly independent set — no vector can be added without creating dependence. Every vector in \( V \) has exactly one coordinate representation relative to a basis.
Why is the coordinate representation unique? Suppose \( u = \sum r_i v_i = \sum s_i v_i \). Then \( \sum (r_i - s_i) v_i = 0 \). Since the \( v_i \) are linearly independent, every coefficient \( r_i - s_i = 0 \), so \( r_i = s_i \).

Dimension of a Vector Space

If \( S = \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \} \) is a basis for a vector space \( V \), then \( n \) — the number of vectors in the basis — is the dimension of \( V \), denoted \( \dim(V) = n \).

Standard basis of \( \mathbb{R}^n \). The standard basis is \( \{ e_1, e_2, \dots, e_n \} \), where \( e_i \) has a \( 1 \) in position \( i \) and \( 0 \) elsewhere. Relative to this basis, the coordinate vector of \( u = (u_1, u_2, \dots, u_n) \) is \( u \) itself.

Definition of Coordinates

If \( S = \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_n \} \) is a basis for a vector space \( V \), and a vector \( \mathbf{u} \in V \) can be written as

\( \mathbf{u} = r_1 \mathbf{v}_1 + r_2 \mathbf{v}_2 + \dots + r_n \mathbf{v}_n \),

then the column vector

\[ [\mathbf{u}]_S = \begin{bmatrix} r_1 \\ r_2 \\ \vdots \\ r_n \end{bmatrix} \]

is called the coordinate vector of \( \mathbf{u} \) relative to the basis \( S \).

Examples with Solutions

Example 1 — Determine Whether a Set Forms a Basis

Is the set of vectors \( S \) a basis for a subspace of \( \mathbb{R}^3 \)?

\[ S = \left\{ \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix}, \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} \right\} \]

Solution

We first check whether the vectors in \( S \) are linearly independent by solving the vector equation

\[ r_1 \begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix} + r_2 \begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} + r_3 \begin{bmatrix} -1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}. \]

Expressed in matrix form, this homogeneous system is:

\[ \begin{bmatrix} -1 & 1 & -1 \\ 1 & 3 & 1 \\ 0 & -1 & 1 \end{bmatrix} \begin{bmatrix} r_1 \\ r_2 \\ r_3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}. \quad \text{(I)} \]

The augmented matrix for this system is:

\[ \begin{bmatrix} -1 & 1 & -1 & | & 0 \\ 1 & 3 & 1 & | & 0 \\ 0 & -1 & 1 & | & 0 \end{bmatrix}. \]

We now row-reduce this augmented matrix to row echelon form:

  • Step 1. Add row 1 to row 2 (i.e., \( R_2 \leftarrow R_2 + R_1 \)): \[ \begin{bmatrix} -1 & 1 & -1 & | & 0 \\ 0 & 4 & 0 & | & 0 \\ 0 & -1 & 1 & | & 0 \end{bmatrix} \]
  • Step 2. Add \( 4 \) times row 3 to row 2 (i.e., \( R_2 \leftarrow R_2 + 4 R_3 \)): \[ \begin{bmatrix} -1 & 1 & -1 & | & 0 \\ 0 & 0 & 4 & | & 0 \\ 0 & -1 & 1 & | & 0 \end{bmatrix} \]
  • Step 3. Interchange rows 2 and 3 (i.e., \( R_2 \leftrightarrow R_3 \)): \[ \begin{bmatrix} -1 & 1 & -1 & | & 0 \\ 0 & -1 & 1 & | & 0 \\ 0 & 0 & 4 & | & 0 \end{bmatrix} \]
  • Step 4. Multiply row 1 by \( -1 \) and row 2 by \( -1 \): \[ \begin{bmatrix} 1 & -1 & 1 & | & 0 \\ 0 & 1 & -1 & | & 0 \\ 0 & 0 & 4 & | & 0 \end{bmatrix} \]
  • Step 5. Multiply row 3 by \( \tfrac{1}{4} \): \[ \begin{bmatrix} 1 & -1 & 1 & | & 0 \\ 0 & 1 & -1 & | & 0 \\ 0 & 0 & 1 & | & 0 \end{bmatrix} \]

Pivots appear in columns 1, 2, and 3. There are no free variables, so the only solution is the trivial one \( r_1 = r_2 = r_3 = 0 \). Hence the vectors in \( S \) are linearly independent.

Now check whether the vectors span \( \mathbb{R}^3 \).

Form the square matrix whose columns are the vectors of \( S \):

\[ M = \begin{bmatrix} -1 & 1 & -1 \\ 1 & 3 & 1 \\ 0 & -1 & 1 \end{bmatrix}. \]

Compute its determinant by cofactor expansion along the first row:

\[ \det(M) = -1 \det\begin{bmatrix} 3 & 1 \\ -1 & 1 \end{bmatrix} - 1 \det\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} + (-1) \det\begin{bmatrix} 1 & 3 \\ 0 & -1 \end{bmatrix} \] \[ = -1(3 + 1) - 1(1 - 0) - 1(-1 - 0) = -4 - 1 + 1 = -4 \neq 0. \]

Since \( \det(M) \neq 0 \), the matrix is invertible, and the columns of \( M \) span \( \mathbb{R}^3 \).

The set \( S \) is both linearly independent and spans \( \mathbb{R}^3 \). Therefore \( S \) is a basis for \( \mathbb{R}^3 \).

Conclusion. \( S \) is a basis for \( \mathbb{R}^3 \), and \( \dim(\mathbb{R}^3) = 3 \) (as expected).
Example 2 — Find Coordinate Vectors Relative to a Basis

Given the basis \( S \) of a subspace \( V \) and a vector \( u \in V \):

\[ S = \left\{ \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 2 \\ 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \\ -1 \\ 1 \end{bmatrix} \right\}, \qquad u = \begin{bmatrix} 4 \\ -3 \\ 0 \\ 1 \end{bmatrix}. \]

(a) What is the dimension of \( V \)? (b) Find the coordinate vector \( [u]_S \).

Solution

(a) The dimension of \( V \) equals the number of vectors in a basis, so \( \dim(V) = 3 \).

(b) Solve \( u = r_1 v_1 + r_2 v_2 + r_3 v_3 \):

\[ \begin{bmatrix} 4 \\ -3 \\ 0 \\ 1 \end{bmatrix} = r_1 \begin{bmatrix} 1 \\ 0 \\ 1 \\ 0 \end{bmatrix} + r_2 \begin{bmatrix} 0 \\ 2 \\ 1 \\ 0 \end{bmatrix} + r_3 \begin{bmatrix} 1 \\ 1 \\ -1 \\ 1 \end{bmatrix}. \]

Augmented matrix:

\[ \begin{bmatrix} 1 & 0 & 1 & | & 4 \\ 0 & 2 & 1 & | & -3 \\ 1 & 1 & -1 & | & 0 \\ 0 & 0 & 1 & | & 1 \end{bmatrix}. \]

Row-reducing gives:

\[ \begin{bmatrix} 1 & 0 & 0 & | & 3 \\ 0 & 1 & 0 & | & -2 \\ 0 & 0 & 1 & | & 1 \\ 0 & 0 & 0 & | & 0 \end{bmatrix}. \]

So \( r_1 = 3 \), \( r_2 = -2 \), \( r_3 = 1 \), and

\[ [u]_S = \begin{bmatrix} 3 \\ -2 \\ 1 \end{bmatrix}. \]

Example 3 — Change of Basis Coordinates

Given the basis \( S_1 \) and coordinate vector \( [u]_{S_1} \):

\[ S_1 = \left\{ \begin{bmatrix} 1 \\ 2 \end{bmatrix}, \begin{bmatrix} -3 \\ 1 \end{bmatrix} \right\}, \qquad [u]_{S_1} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}. \]

Find \( [u]_{S_2} \) relative to the new basis \( S_2 = \left\{ \begin{bmatrix} 0 \\ -1 \end{bmatrix}, \begin{bmatrix} -1 \\ 2 \end{bmatrix} \right\} \).

Solution

First, recover the standard vector representation:

\[ u = -1 \begin{bmatrix} 1 \\ 2 \end{bmatrix} + 2 \begin{bmatrix} -3 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 - 6 \\ -2 + 2 \end{bmatrix} = \begin{bmatrix} -7 \\ 0 \end{bmatrix}. \]

Now solve \( u = r_1 w_1 + r_2 w_2 \) for \( S_2 \):

\[ \begin{bmatrix} -7 \\ 0 \end{bmatrix} = r_1 \begin{bmatrix} 0 \\ -1 \end{bmatrix} + r_2 \begin{bmatrix} -1 \\ 2 \end{bmatrix}. \]

Augmented matrix:

\[ \begin{bmatrix} 0 & -1 & | & -7 \\ -1 & 2 & | & 0 \end{bmatrix}. \]

Row-reducing gives:

\[ \begin{bmatrix} 1 & -2 & | & 0 \\ 0 & 1 & | & 7 \end{bmatrix}. \]

Back-substituting: \( r_2 = 7 \) and \( r_1 = 14 \). Therefore

\[ [u]_{S_2} = \begin{bmatrix} 14 \\ 7 \end{bmatrix}. \]

Example 4 — Find a Basis for a Spanned Subspace

Find a basis for the subspace of \( \mathbb{R}^3 \) spanned by:

\[ \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \; \begin{bmatrix} -1 \\ -2 \\ -7 \end{bmatrix}, \; \begin{bmatrix} 1 \\ 0 \\ -3 \end{bmatrix}, \; \begin{bmatrix} 2 \\ 0 \\ 6 \end{bmatrix}. \]

Solution

Place the vectors as columns of a matrix and row-reduce:

\[ A = \begin{bmatrix} 1 & -1 & 1 & 2 \\ 1 & -2 & 0 & 0 \\ 2 & -7 & -3 & 6 \end{bmatrix} \xrightarrow{\;\text{RREF}\;}\; \begin{bmatrix} 1 & 0 & 2 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}. \]

Pivot columns are 1, 2, and 4. The corresponding columns of the original matrix form a basis for the column space:

\[ \left\{ \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}, \begin{bmatrix} -1 \\ -2 \\ -7 \end{bmatrix}, \begin{bmatrix} 2 \\ 0 \\ 6 \end{bmatrix} \right\}. \]

Note: the dimension of the column space equals the number of pivots in the row echelon form.

Questions with Solutions

Question 1 — Is It a Basis?

Is \( S = \left\{ \begin{bmatrix} 1 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right\} \) a basis for \( \mathbb{R}^2 \)? Justify your answer.

Solution

No. A basis for \( \mathbb{R}^2 \) must have exactly 2 vectors. The set \( S \) has 3, so it cannot be a basis. In fact, the three vectors are linearly dependent: \( (1,1) = (1,0) + (0,1) \).

Question 2 — Find Coordinates Relative to a Basis

Find the coordinate vector of \( u = \begin{bmatrix} 5 \\ 6 \end{bmatrix} \) relative to \( S = \left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ -1 \end{bmatrix} \right\} \).

Solution

Solve \( r_1 (1,1) + r_2 (1,-1) = (5,6) \):

\[ \begin{cases} r_1 + r_2 = 5 \\ r_1 - r_2 = 6 \end{cases} \]

Adding: \( 2r_1 = 11 \Rightarrow r_1 = \tfrac{11}{2} \). Subtracting: \( 2r_2 = -1 \Rightarrow r_2 = -\tfrac{1}{2} \).

\[ [u]_S = \begin{bmatrix} 11/2 \\ -1/2 \end{bmatrix}. \]

Question 3 — Dimension of a Column Space

Find a basis for the column space of \( A = \begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 1 \\ 1 & 3 & 1 \end{bmatrix} \), and state its dimension.

Solution

Row-reduce \( A \):

\[ A \xrightarrow{\;\text{RREF}\;}\; \begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}. \]

Pivot columns are 1 and 2. A basis for the column space is:

\[ \left\{ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} \right\}. \]

The dimension of the column space (the rank of \( A \)) is \( 2 \).

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