The Ultimate Guide to Understanding Basis and Dimension in Vector Spaces

How to Determine Whether a Set Forms a Basis

To determine whether a set forms a basis, we must verify both span and linear independence.

The exact procedure depends on the vector space involved, but the general strategy is always the same.

Suppose we are given vectors

$${v_1,v_2,\dots,v_n}.$$

To show that these form a basis, we proceed as follows.

Step 1: Check Linear Independence

Determine whether the equation

$$c_1v_1 + c_2v_2 + \cdots + c_nv_n = 0$$

has only the trivial solution

$$c_1 = c_2 = \cdots = c_n = 0.$$

If the only solution is the trivial solution, the vectors are linearly independent.

This often involves solving a homogeneous system of equations using Gaussian elimination. See How to Solve a System of Equations Using Gaussian Elimination: A Step-by-Step Guide for a detailed review of this process.

It may also require computing a determinant. For a review of determinants, please refer to How to Find the Determinant of a Matrix Step by Step: A Complete Beginner’s Guide.

Step 2: Check Span

Determine whether every vector in the space can be written as a linear combination of the given vectors.

For a review of linear combinations, please refer to the articles The Ultimate Step-by-Step Guide to Basic Matrix Operations for Beginners and Vector Operations Tutorial for Beginners: A Complete Step-by-Step Guide.

Step 3: Conclude Whether the Set Is a Basis

If the vectors are both linearly independent and form a spanning set, they form a basis. If either condition fails, the set is not a basis.

Basis Examples

Examples in \( \mathbb{R}^n \)

Example 1: Prove or disprove:
(a) \( \{ \langle 1, 2 \rangle, \langle 3, 1 \rangle \} \) forms a basis for \( \mathbb{R}^2 \).
(b) \( \{ \langle 1, 0, 2 \rangle, \langle 2, 1, 3 \rangle, \langle 0, 1, 1 \rangle \} \) forms a basis for \( \mathbb{R}^3 \).
(c) \( \{ \langle 1, 2 \rangle, \langle 2, 4 \rangle \} \) forms a basis for \( \mathbb{R}^2 \).

Solution: (a) \( \{ \langle 1, 2 \rangle, \langle 3, 1 \rangle \} \) forms a basis for \( \mathbb{R}^2 \).

From the definition of linear independence/dependence, we have

$$c_1\langle 1, 2 \rangle + c_2\langle 3, 1 \rangle = 0.$$

Multiplying each component of the first vector by \( c_1 \) and each component of the second vector by \( c_2 \), we get

$$\langle c_1, 2c_1 \rangle + \langle 3c_2, c_2 \rangle = 0.$$

Adding corresponding components gives

$$\langle c_1 + 3c_2, 2c_1 + c_2 \rangle = 0.$$

Next, we equate components to obtain the system of equations

$$\begin{aligned}
c_1 + 3c_2 &= 0 \\
2c_1 + c_2 &= 0
\end{aligned}.$$

In matrix form, this is

$$\begin{bmatrix}
1 & 3 \\
2 & 1
\end{bmatrix}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0
\end{bmatrix}.$$

We now take the determinant of the coefficient matrix to get

$$\begin{vmatrix}
1 & 3 \\
2 & 1
\end{vmatrix}.$$

Applying the formula for 2×2 determinants gives

$$(1)(1) – (3)(2).$$

Simplifying we get

$$1 – 6.$$

Subtracting, we arrive at

$$-5.$$

Since the determinant is nonzero, the only solution is the trivial solution, so the vectors are linearly independent. The span is given by

$$c_1\langle 1, 2 \rangle + c_2\langle 3, 1 \rangle.$$

Multiplying each component of the first vector by \( c_1 \) and each component of the second vector by \( c_2 \), we get

$$\langle c_1, 2c_1 \rangle + \langle 3c_2, c_2 \rangle.$$

Adding corresponding components gives

$$\langle c_1 + 3c_2, 2c_1 + c_2 \rangle.$$

Since \( c_1 \) and \( c_2 \) can be any real numbers, the span is

$$\mathbb{R}^2.$$

Since the vectors are linearly independent and span \( \mathbb{R}^2 \), the set forms a basis for \( \mathbb{R}^2 \).

(b) \( \{ \langle 1, 0, 2 \rangle, \langle 2, 1, 3 \rangle, \langle 0, 1, 1 \rangle \} \) forms a basis for \( \mathbb{R}^3 \).

From the definition of linear independence/dependence, we have

$$c_1\langle 1, 0, 2 \rangle + c_2\langle 2, 1, 3 \rangle + c_3\langle 0, 1, 1 \rangle = 0.$$

Multiplying each component of the first vector by \( c_1 \), each component of the second vector by \( c_2 \), and each component of the third vector by \( c_3 \), we get

$$\langle c_1, 0, 2c_1 \rangle + \langle 2c_2, c_2, 3c_2 \rangle + \langle 0, c_3, c_3 \rangle = 0.$$

Adding corresponding components gives

$$\langle c_1 + 2c_2, c_2 + c_3, 2c_1 + 3c_2 + c_3 \rangle = 0.$$

Next, we equate components to obtain the system of equations

$$\begin{aligned}
c_1 + 2c_2 &= 0 \\
c_2 + c_3 &= 0 \\
2c_1 + 3c_2 + c_3 &= 0
\end{aligned}.$$

In matrix form, this is

$$\begin{bmatrix}
1 & 2 & 0 \\
0 & 1 & 1 \\
2 & 3 & 1
\end{bmatrix}
\begin{bmatrix}
c_1 \\
c_2 \\
c_3
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.$$

We now take the determinant of the coefficient matrix to get

$$\begin{vmatrix}
1 & 2 & 0 \\
0 & 1 & 1 \\
2 & 3 & 1
\end{vmatrix}.$$

Extending the matrix gives

$$\begin{vmatrix}
1 & 2 & 0 \\
0 & 1 & 1 \\
2 & 3 & 1
\end{vmatrix}
\begin{matrix}
1 & 2 \\
0 & 1 \\
2 & 3
\end{matrix}.$$

Computing the determinant by summing the products of the three diagonals running from the top-left to the bottom-right and subtracting the products of the three diagonals running from the bottom-left to the top-right, we obtain

$$(1)(1)(1) + (2)(1)(2) + (0)(0)(3) – (2)(1)(0) – (3)(1)(1) – (1)(0)(2).$$

Simplifying we get

$$1 + 4 + 0 – 0 – 3 – 0.$$

Adding and subtracting, we arrive at

$$2.$$

Since the determinant is nonzero, the only solution is the trivial solution, so the vectors are linearly independent. The span is given by

$$c_1\langle 1, 0, 2 \rangle + c_2\langle 2, 1, 3 \rangle + c_3\langle 0, 1, 1 \rangle.$$

Multiplying each component of the first vector by \( c_1 \), each component of the second vector by \( c_2 \), and each component of the third vector by \( c_3 \), we get

$$\langle c_1, 0, 2c_1 \rangle + \langle 2c_2, c_2, 3c_2 \rangle + \langle 0, c_3, c_3 \rangle.$$

Adding corresponding components gives

$$\langle c_1 + 2c_2, c_2 + c_3, 2c_1 + 3c_2 + c_3 \rangle.$$

Since \( c_1 \), \( c_2 \), and \( c_3 \) can be any real numbers, the span is

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

Since the vectors are linearly independent and span \( \mathbb{R}^3 \), the set forms a basis for \( \mathbb{R}^3 \).

(c) \( \{ \langle 1, 2 \rangle, \langle 2, 4 \rangle \} \) forms a basis for \( \mathbb{R}^2 \).

From the definition of linear independence/dependence, we have

$$c_1 \langle 1, 2 \rangle + c_2 \langle 2, 4 \rangle = 0.$$

Multiplying each component of the first vector by \( c_1 \) and each component of the second vector by \( c_2 \), we get

$$\langle c_1, 2c_1 \rangle + \langle 2c_2, 4c_2 \rangle = 0.$$

Adding corresponding components gives

$$\langle c_1 + 2c_2, 2c_1 + 4c_2 \rangle = 0.$$

Next, we equate components to obtain the system of equations

$$\begin{aligned}
c_1 + 2c_2 &= 0 \\
2c_1 + 4c_2 &= 0
\end{aligned},$$

In matrix form, this is

$$\begin{bmatrix}
1 & 2 \\
2 & 4
\end{bmatrix}
\begin{bmatrix}
c_1 \\
c_2
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0
\end{bmatrix}.$$

We now take the determinant of the coefficient matrix to get

$$\begin{vmatrix}
1 & 2 \\
2 & 4
\end{vmatrix}.$$

Applying the formula for 2×2 determinants gives

$$(1)(4) – (2)(2).$$

Simplifying we get

$$4 – 4.$$

Subtracting, we arrive at

$$0.$$

Since the determinant is 0, the system has nontrivial solutions, so the set is linearly dependent, therefore it is not a basis.

Examples with Matrices

Example 2: Prove or disprove:

(a) $$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \}$$ forms a basis for \( M_{2 \times 2} \).
(b) $$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} \}$$ forms a basis for \( M_{2 \times 2} \).

Solution: (a) $$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \}$$ forms a basis for \( M_{2 \times 2} \).

From the definition of linear independence/dependence, we have

$$c_1\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + c_2\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + c_3\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + c_4\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = 0.$$

Multiplying each entry of the first matrix by \( c_1 \), each entry of the second matrix by \( c_2 \), each entry of the third matrix by \( c_3 \), and each entry of the fourth matrix by \( c_4 \), we get

$$\begin{bmatrix} c_1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & c_2 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ c_3 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & c_4 \end{bmatrix} = 0.$$

Adding corresponding entries, we get

$$\begin{bmatrix} c_1 & c_2 \\ c_3 & c_4 \end{bmatrix} = 0.$$

Next, we equate entries to obtain the system of equations

$$\begin{aligned}
c_1 = 0 \\
c_2 = 0 \\
c_3 = 0 \\
c_4 = 0
\end{aligned}.$$

Thus, the solution is

$$c_1 = 0, c_2 = 0, c_3 = 0, c_4 = 0.$$

The only solution is the trivial solution, so the vectors are linearly independent. The span is given by

$$c_1\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + c_2\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} + c_3\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} + c_4\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}.$$

Multiplying each entry of the first matrix by \( c_1 \), each entry of the second matrix by \( c_2 \), each entry of the third matrix by \( c_3 \), and each entry of the fourth matrix by \( c_4 \), we get

$$\begin{bmatrix} c_1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & c_2 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ c_3 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & c_4 \end{bmatrix}.$$

Adding corresponding entries, we get

$$\begin{bmatrix} c_1 & c_2 \\ c_3 & c_4 \end{bmatrix}.$$

Since \( c_1 \), \( c_2 \), \( c_3 \), and \( c_4 \) can be any real numbers, the span is

$$M_{2 \times 2}.$$

Since the vectors are linear independent and span \( M_{2 \times 2} \), the set forms a basis for \( M_{2 \times 2} \).

(b) $$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} \}$$ forms a basis for \( M_{2 \times 2} \).

From the definition of linear independence/dependence, we have

$$c_1\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + c_2\begin{bmatrix} 2 & 0 \\ 0 & 0 \end{bmatrix} = 0.$$

Multiplying each entry of the first matrix by \( c_1 \) and each entry of the second matrix by \( c_2 \), we get

$$\begin{bmatrix} c_1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 2c_2 & 0 \\ 0 & 0 \end{bmatrix} = 0.$$

Adding corresponding entries, we get

$$\begin{bmatrix} c_1 + 2c_2 & 0 \\ 0 & 0\end{bmatrix} = 0.$$

Next, we equate entries to obtain the system of equations

$$\begin{aligned}
c_1 + 2c_2 &= 0 \\
0 &= 0 \\
0 &= 0 \\
0 &= 0
\end{aligned}.$$

The last three equations do not add anything new, hence this system is equivalent to

$$\begin{aligned}
c_1 + 2c_2 &= 0
\end{aligned}.$$

Which implies

$$\begin{aligned}
c_1 &= -2c_2
\end{aligned}.$$

Letting \( c_2 =t \) be the free variable, our solution is

$$c_1 = -2t, c_2 = t.$$

The system has nontrivial solutions, so the set is linearly dependent, therefore it is not a basis.

Examples with Polynomials

Example 3: Prove or disprove:
(a) \( \{ 1, x, x^2 \} \) forms a basis for \( P_2 \).
(b) \( \{ 1+x, 1 – x, x^2 \} \) forms a basis for \( P_2 \).
(c) \( \{ 1 + x, 2 + 2x \} \) forms a basis for \( P_1 \).

Solution: (a) \( \{ 1, x, x^2 \} \) forms a basis for \( P_2 \).

From the definition of linear independence/dependence, we have

$$c_1(1) + c_2(x) + c_3(x^2) = 0.$$

Distributing, we obtain

$$c_1 + c_2x + c_3x^2 = 0.$$

Next, we equate coefficients to obtain the solution

$$c_1 = 0, c_2 = 0, c_3 = 0.$$

The only solution is the trivial solution, so the vectors are linearly independent. The linear combination is

$$c_1(1) + c_2(x) + c_3(x^2)$$

Distributing, we obtain

$$c_1 + c_2x + c_3x^2$$

Since \(c_1\), \(c_2\), and \(c_3\) can be any real numbers, the span is

$$P_2.$$

Since the vectors are linearly independent and span \( P_2 \), the set forms a basis for \( P_2 \).

(b) \( \{ 1+x, 1 – x, x^2 \} \) forms a basis for \( P_2 \).

From the definition of linear independence/dependence, we have

$$c_1(1 + x) + c_2(1 – x) + c_3x^2 = 0.$$

Distributing, we obtain

$$c_1 + c_1x + c_2 – c_2x + c_3x^2 = 0.$$

Adding, we get

$$(c_1 + c_2) + (c_1 – c_2)x + c_3x^2 = 0.$$

Next, we equate coefficients to obtain the system of equations

$$\begin{aligned}
c_1 + c_2 &= 0 \\
c_1 – c_2 &= 0 \\
c_3 &= 0
\end{aligned}.$$

In matrix form, this is

$$\begin{bmatrix}
1 & 1 & 0 \\
1 & -1 & 0 \\
0 & 0 & 1
\end{bmatrix}
\begin{bmatrix}
c_1 \\
c_2 \\
c_3
\end{bmatrix}
=
\begin{bmatrix}
0 \\
0 \\
0
\end{bmatrix}.$$

We now take the determinant of the coefficient matrix to get

$$\begin{vmatrix}
1 & 1 & 0 \\
1 & -1 & 0 \\
0 & 0 & 1
\end{vmatrix}.$$

Extending the matrix gives

$$\begin{vmatrix}
1 & 1 & 0 \\
1 & -1 & 0 \\
0 & 0 & 1
\end{vmatrix}
\begin{matrix}
1 & 1 \\
1 & -1 \\
0 & 0
\end{matrix}.$$

Computing the determinant by summing the products of the three diagonals running from the top-left to the bottom-right and subtracting the products of the three diagonals running from the bottom-left to the top-right, we obtain

$$(1)(-1)(1) + (1)(0)(0) + (0)(1)(0) – (0)(-1)(0) – (0)(0)(1) – (1)(1)(1).$$

Simplifying we get

$$-1 + 0 + 0 – 0 – 0 – 1.$$

Adding and subtracting, we arrive at

$$-2.$$

Since the determinant is nonzero, the only solution is the trivial solution, so the vectors are linearly independent. The span is given by

$$c_1(1 + x) + c_2(1 – x) + c_3x^2.$$

Distributing, we obtain

$$c_1 + c_1x + c_2 – c_2x + c_3x^2.$$

Adding, we get

$$(c_1 + c_2) + (c_1 – c_2)x + c_3x^2.$$

Since \( c_1 \), \( c_2 \), and \( c_3 \) can be any real numbers, the span is

$$P_2.$$

Since the vectors are linearly independent and span \( P_2 \), the set forms a basis for \( P_2 \).

(c) \( \{ 1 + x, 2 + 2x \} \) forms a basis for \( P_1 \).

From the definition of linear independence/dependence, we have

$$c_1(1 + x) + c_2(2 + 2x) = 0.$$

Distributing, we obtain

$$c_1 + c_1x + 2c_2 + 2c_2x = 0.$$

Adding, we get

$$(c_1 + 2c_2) + (c_1 + 2c_2)x = 0.$$

Next, we equate coefficients to obtain the system of equations

$$\begin{aligned}
c_1 + 2c_2 &= 0 \\
c_1 + 2c_2 &= 0
\end{aligned}.$$

The last equation adds nothing new, hence this system is equivalent to

$$\begin{aligned}
c_1 + 2c_2 &= 0
\end{aligned}.$$

Which implies

$$\begin{aligned}
c_1 &= -2c_2
\end{aligned}.$$

Letting \( c_2 =t \) be the free variable, our solution is

$$c_1 = -2t, c_2 = t.$$

The system has nontrivial solutions, so the set is linearly dependent, therefore it is not a basis.

Dimension of a Vector Space

Once we understand the concept of basis, the next important idea is dimension. Dimension tells us how large a vector space is by measuring the number of vectors required in a basis.

Definition

The dimension of a vector space \( V \) is the number of vectors in any basis for \( V \) and is denoted by

$$\dim(V) = n.$$

Finite and Infinite Dimensional Spaces

A vector space is finite-dimensional if it has a finite basis, and infinite-dimensional if no finite basis exists.

How to Compute Dimension

Note that:

• \( \dim(\mathbb{R}^n) = n \).
• \( \dim(M_{m \times n}) = mn \).
• \( \dim(P_n) = n + 1 \).
• If \( W \subset V \), then \( \dim(W) \leq \dim(V) \).

Relationship Between Basis and Dimension

Every basis for the same vector space contains the same number of vectors. For example, every basis for \( \mathbb{R}^2 \) must contain exactly two vectors. This guarantees the dimension is well-defined.

Examples

Example 4: Compute the dimension of the following vector spaces:
(a) \( \mathbb{R}^2 \)
(b) \( \mathbb{R}^3 \)
(c) \( M_{2 \times 2} \)
(d) \( P_2 \)

Solution: (a) \( \mathbb{R}^2 \)

Applying the formula, we arrive at a final answer of

$$2.$$

(b) \( \mathbb{R}^3 \)

Applying the formula, we arrive at a final answer of

$$3.$$

(c) \( M_{2 \times 2} \)

Applying the formula gives

$$2(2).$$

Multiplying, we arrive at a final answer of

$$4.$$

(d) \( P_2 \)

Applying the formula gives

$$2 + 1.$$

Adding, we arrive at a final answer of

$$3.$$

How to Expand a Linearly Independent Set into a Basis

Sometimes a set of vectors is linearly independent but does not span the entire vector space. In this case, the set is too small to form a basis. To fix this, we add additional vectors to the set that are not in the span, while preserving linear independence.

Examples

Example 5: Expand the following linearly independent sets into a basis for the given vector space:
(a) \( \{ \langle 1, 2 \rangle \} \), \( \mathbb{R}^2 \)
(b) \( \{ \langle 1, 0, 0 \rangle, \langle 0, 1, 0 \rangle \} \), \( \mathbb{R}^3 \)
(c) $$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \}, M_{2 \times 2}$$
(d) \( \{ 1 + x \} \), \( P_2 \).

Solution: (a) \( \{ \langle 1, 2 \rangle \} \), \( \mathbb{R}^2 \).

Notice that \( \langle 0, 1 \rangle \notin \text{span}\{ \langle 1, 2 \rangle \} \), hence a final basis is

$$\{ \langle 1, 2 \rangle, \langle 0,1 \rangle \}.$$

(b) \( \{ \langle 1, 0, 0 \rangle, \langle 0, 1, 0 \rangle \} \), \( \mathbb{R}^3 \).

Notice that $$\langle 0, 0, 1 \rangle \notin \text{span}\{ \langle 1, 0, 0 \rangle, \langle 0, 1, 0 \rangle \},$$ hence a final basis is

$$\{ \langle 1, 0, 0 \rangle, \langle 0, 1, 0 \rangle, \langle 0, 0, 1 \rangle \}.$$

(c) $$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \}, M_{2 \times 2}.$$

Notice that \( \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \notin \text{span}\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \} \), expanding the set then gives

$$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \}.$$

Notice that $$\begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}\notin \text{span}\{\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \},$$ expanding the set once again we get

$$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \}.$$

Notice that $$\begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \notin \text{span}\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} \},$$

hence a final basis is

$$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \}.$$

(d) \( \{ 1 + x \} \), \( P_2 \).

Notice that \( x \notin \text{span}\{ 1 + x \} \), expanding the set then gives

$$\{ 1 + x, x \}.$$

Notice that \( x^2 \notin \text{span}\{ 1 + x, x \} \), hence a final basis is

$$\{ 1 + x, x, x^2 \}.$$

How to Shrink a Spanning Set into a Basis

Sometimes a set spans a vector space but contains too many vectors, making it linearly dependent. To create a basis, we remove unnecessary vectors while preserving span.

Examples

Example 6: Shrink the following spanning sets into a basis for the given vector space:
(a) \( \{ \langle1, 0 \rangle, \langle 0, 1 \rangle, \langle 1, 1 \rangle \} \), \( \mathbb{R}^2 \)
(b) $$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \}, M_{2 \times 2}$$
(c) \( \{ 1, x, 1+x, x^2 \} \), \( P_2 \).

Solution: (a) \( \{ \langle1, 0 \rangle, \langle 0, 1 \rangle, \langle 1, 1 \rangle \} \), \( \mathbb{R}^2 \)

Notice that \( \langle1, 0 \rangle + \langle 0, 1 \rangle = \langle 1, 1 \rangle \), hence a final basis is

$$\{ \langle1, 0 \rangle, \langle 0, 1 \rangle \}.$$

(b) $$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix}, \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \}, M_{2 \times 2} \}$$

Notice that $$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix},$$ hence a final basis is

$$\{ \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix}, \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} \}.$$

(c) \( \{ 1, x, 1+x, x^2 \} \), \( P_2 \).

Notice that \( 1 + x = 1 + x \), hence a final basis is

$$\{ 1, x, x^2 \}.$$

Conclusion

Understanding basis and dimension in vector spaces is essential because these concepts reveal how vector spaces are organized and how vectors interact within them. A basis provides the fundamental building blocks of a vector space, while dimension tells us how many independent vectors are needed to describe the entire space.

In this guide, we explored how to determine whether a set forms a basis by checking span and linear independence. We studied how dimension is computed and how it relates directly to the number of vectors in a basis. We also learned how to expand linearly independent sets into bases and how to shrink spanning sets by removing redundant vectors. Along the way, we worked through examples involving vectors in \( \mathbb{R}^n \), matrices, and polynomial vector spaces.

One of the most important ideas to remember is that basis and dimension work together. A basis must contain enough vectors to span the space, but not so many that redundancy appears. Dimension then measures exactly how many vectors are required.

Further Reading

A Comprehensive Beginner’s Guide to Partial Fraction Decomposition – With your knowledge of basis, I recommend revisiting partial fraction decomposition and testing whether the set of fractions in the decomposition forms a basis.

Step-by-Step Solutions to Common Examples of Linear Transformations in Linear Algebra – It is also informative to revisit linear transformations and explore if the columns of the transformation matrices form a basis.

Frequently Asked Questions