How To Find Basis Of Subspace

Finding the basis of a subspace is a fundamental concept in linear algebra, crucial for understanding vector spaces and their properties. A basis is a set of linearly independent vectors that span the entire subspace. This means that any vector in the subspace can be expressed as a linear combination of the basis vectors, and none of the basis vectors can be written as a linear combination of the others. This article delves into the process of finding the basis of a subspace, providing detailed explanations and practical examples.

Understanding Subspaces and Bases

Before diving into the methods, let's clarify what subspaces and bases are.

• Subspace: A subspace V of a vector space W is a subset of W that is itself a vector space. This means it must satisfy three conditions:

  1. The zero vector of W is in V.
  2. If u and v are in V, then u + v is in V (closed under addition).
  3. If u is in V and c is a scalar, then cu is in V (closed under scalar multiplication).

• Basis: A basis for a subspace V is a set of vectors {v₁, v₂, ..., v_n} in V that satisfies two conditions:

  1. The vectors are linearly independent.
  2. The vectors span V.

• Linear Independence: A set of vectors is linearly independent if no vector in the set can be written as a linear combination of the other vectors. In other words, the only solution to the equation c₁v₁ + c₂v₂ + ... + c_nv_n = 0 is c₁ = c₂ = ... = c_n = 0.

• Spanning: A set of vectors {v₁, v₂, ..., v_n} spans a subspace V if every vector in V can be written as a linear combination of the vectors in the set. That is, for any vector v in V, there exist scalars c₁, c₂, ..., c_n such that v = c₁v₁ + c₂v₂ + ... + c_nv_n.

Methods for Finding the Basis of a Subspace

Several methods can be employed to find the basis of a subspace, depending on how the subspace is defined. Here are some common approaches:

1. Subspace Defined by a Spanning Set:
   • Method: Reduction to Linear Independence
   • Description: If you are given a set of vectors that spans a subspace, the goal is to reduce this set to a linearly independent set that still spans the same subspace. This is typically done by identifying and removing linearly dependent vectors.
2. Subspace Defined by Equations:
   • Method: Solving Homogeneous Systems of Linear Equations
   • Description: When a subspace is defined by a set of linear equations, you can find the basis by solving the corresponding homogeneous system of equations. The solutions to this system form the subspace, and you can express these solutions in terms of free variables, which then give you the basis vectors.
3. Column Space of a Matrix:
   • Method: Finding Pivot Columns
   • Description: The column space of a matrix A is the span of the columns of A. To find the basis for the column space, you can perform row reduction to echelon form. The columns of the original matrix A corresponding to the pivot columns in the echelon form constitute a basis for the column space.
4. Null Space of a Matrix:
   • Method: Finding Free Variables and Expressing Solutions
   • Description: The null space of a matrix A is the set of all vectors x such that Ax = 0. To find the basis for the null space, you solve the homogeneous equation Ax = 0. The solutions can be expressed in terms of free variables, and the coefficients of these free variables form the basis vectors for the null space.
5. Eigenspace of a Matrix:
   • Method: Solving (A - λI)x = 0
   • Description: For a given eigenvalue λ of a matrix A, the eigenspace is the set of all eigenvectors corresponding to that eigenvalue. This is found by solving the equation (A - λI)x = 0, where I is the identity matrix. The basis of this eigenspace is then the set of linearly independent eigenvectors.

Detailed Steps and Examples

Let's explore each method with detailed steps and examples to illustrate the process.

1. Subspace Defined by a Spanning Set: Reduction to Linear Independence

Steps:

1. Write the vectors as columns of a matrix: Given a set of vectors that span a subspace, create a matrix A with these vectors as its columns.
2. Perform row reduction to echelon form: Apply Gaussian elimination or other row reduction techniques to transform the matrix A into its row echelon form.
3. Identify pivot columns: Determine the pivot columns, which are the columns containing the leading entries (pivots) in the echelon form.
4. The original vectors corresponding to the pivot columns form the basis: The vectors in the original matrix A that correspond to the pivot columns in the echelon form are the basis vectors for the subspace.

Example:

Suppose the subspace V of R⁴ is spanned by the vectors:

v₁ = (1, 2, 3, 1) v₂ = (2, 4, 6, 2) v₃ = (1, 3, 4, 2) v₄ = (1, 1, 1, 0)

Solution:

1. Form the matrix A:

   A = | 1  2  1  1 |
       | 2  4  3  1 |
       | 3  6  4  1 |
       | 1  2  2  0 |
   

2. Row reduce to echelon form:

   Applying row operations:

   • R2 -> R2 - 2R1
   • R3 -> R3 - 3R1
   • R4 -> R4 - R1

   A = | 1  2  1  1 |
       | 0  0  1 -1 |
       | 0  0  1 -2 |
       | 0  0  1 -1 |
   

   • R3 -> R3 - R2
   • R4 -> R4 - R2

   A = | 1  2  1  1 |
       | 0  0  1 -1 |
       | 0  0  0 -1 |
       | 0  0  0  0 |
   

   • R3 -> -R3

   A = | 1  2  1  1 |
       | 0  0  1 -1 |
       | 0  0  0  1 |
       | 0  0  0  0 |
   

3. Identify pivot columns: The pivot columns are the first, third, and fourth columns.

4. Basis vectors: The basis vectors for the subspace V are the original vectors corresponding to the pivot columns:

   v₁ = (1, 2, 3, 1) v₃ = (1, 3, 4, 2) v₄ = (1, 1, 1, 0)

Therefore, the basis for V is {v₁, v₃, v₄}.

2. Subspace Defined by Equations: Solving Homogeneous Systems of Linear Equations

Steps:

1. Write the equations in matrix form: Express the given equations as a homogeneous system of linear equations in the form Ax = 0.
2. Solve the system: Find the general solution to the system Ax = 0 by row reducing the augmented matrix [A | 0] to reduced row echelon form.
3. Express the solution in terms of free variables: Identify the free variables and express the basic variables in terms of the free variables.
4. Write the general solution as a linear combination of vectors: Express the general solution as a linear combination of vectors, where the coefficients are the free variables.
5. The vectors in the linear combination form the basis: The vectors that multiply the free variables form the basis for the subspace.

Example:

Suppose the subspace V of R³ is defined by the equation:

x + y + z = 0

Solution:

1. Matrix form: The equation can be written as Ax = 0, where

   A = | 1  1  1 |
   x = | x |
       | y |
       | z |
   

2. Solve the system: The augmented matrix is [1 1 1 | 0]. This matrix is already in reduced row echelon form.

3. Express the solution in terms of free variables: Let y = s and z = t, where s and t are free variables. Then, x = -s - t.

4. Write the general solution as a linear combination: The general solution can be written as:

   | x |   | -s - t |   | -s |   | -t |   s | -1 |   t | -1 |
   | y | = |    s   | = |  s | + |  0 | = s |  1 | + t |  0 |
   | z |   |    t   |   |  0 |   |  t |   s |  0 |   t |  1 |
   

5. Basis vectors: The basis vectors for the subspace V are:

   v₁ = (-1, 1, 0) v₂ = (-1, 0, 1)

Therefore, the basis for V is {(-1, 1, 0), (-1, 0, 1)}.

3. Column Space of a Matrix: Finding Pivot Columns

Steps:

1. Form the matrix: Given a matrix A, the column space is the span of its column vectors.
2. Row reduce to echelon form: Apply row operations to transform the matrix A into its row echelon form.
3. Identify pivot columns: Determine the pivot columns in the echelon form.
4. The original columns corresponding to the pivot columns form the basis: The corresponding columns in the original matrix A form the basis for the column space.

Example:

Consider the matrix:

A = | 1  2  1 |
    | 2  4  2 |
    | 3  6  3 |

Solution:

1. Form the matrix: (already given)

2. Row reduce to echelon form:

   • R2 -> R2 - 2R1
   • R3 -> R3 - 3R1

   A = | 1  2  1 |
       | 0  0  0 |
       | 0  0  0 |
   

3. Identify pivot columns: The first column is the only pivot column.

4. Basis vector: The first column of the original matrix A is the basis vector:

   v₁ = (1, 2, 3)

Therefore, the basis for the column space of A is {(1, 2, 3)}.

4. Null Space of a Matrix: Finding Free Variables and Expressing Solutions

Steps:

1. Form the matrix: Given a matrix A, the null space is the set of all vectors x such that Ax = 0.
2. Solve the homogeneous equation Ax** = 0: Row reduce the augmented matrix [A | 0] to reduced row echelon form.
3. Express the solution in terms of free variables: Identify the free variables and express the basic variables in terms of the free variables.
4. Write the general solution as a linear combination of vectors: Express the general solution as a linear combination of vectors, where the coefficients are the free variables.
5. The vectors in the linear combination form the basis: The vectors that multiply the free variables form the basis for the null space.

Example:

Consider the matrix:

A = | 1  2  1 |
    | 2  4  2 |
    | 3  6  3 |

Solution:

1. Form the matrix: (already given)

2. Solve Ax** = 0: Row reduce the augmented matrix [A | 0]:

   | 1  2  1 | 0 |
   | 2  4  2 | 0 |
   | 3  6  3 | 0 |
   

   • R2 -> R2 - 2R1
   • R3 -> R3 - 3R1

   | 1  2  1 | 0 |
   | 0  0  0 | 0 |
   | 0  0  0 | 0 |
   

3. Express the solution in terms of free variables: Let x₂ = s and x₃ = t. Then x₁ = -2s - t.

4. Write the general solution as a linear combination:

   | x1 |   | -2s - t |   | -2s |   | -t |   s | -2 |   t | -1 |
   | x2 | = |    s    | = |  s  | + |  0 | = s |  1 | + t |  0 |
   | x3 |   |    t    |   |  0  |   |  t |   s |  0 |   t |  1 |
   

5. Basis vectors: The basis vectors for the null space of A are:

   v₁ = (-2, 1, 0) v₂ = (-1, 0, 1)

Therefore, the basis for the null space of A is {(-2, 1, 0), (-1, 0, 1)}.

5. Eigenspace of a Matrix: Solving (A - λI)x = 0

Steps:

1. Find the eigenvalues: Determine the eigenvalues λ of the matrix A by solving the characteristic equation det(A - λI) = 0.
2. For each eigenvalue, solve (A - λI)x = 0: For each eigenvalue λ, solve the homogeneous system (A - λI)x = 0. This gives the eigenvectors corresponding to that eigenvalue.
3. Express the solution in terms of free variables: Identify the free variables and express the basic variables in terms of the free variables.
4. Write the general solution as a linear combination of vectors: Express the general solution as a linear combination of vectors, where the coefficients are the free variables.
5. The vectors in the linear combination form the basis: The vectors that multiply the free variables form the basis for the eigenspace corresponding to the eigenvalue λ.

Example:

Consider the matrix:

A = | 2  0 |
    | 0  3 |

Solution:

1. Find the eigenvalues:

   det(A - λI) = det(| 2-λ 0 |) = (2-λ)(3-λ) - 0*0 = (2-λ)(3-λ) = 0 (| 0 3-λ |)

   λ₁ = 2, λ₂ = 3

2. Solve (A - λI)x = 0 for each eigenvalue:

   • For λ₁ = 2:

     (A - 2I)x = 0

     | 0  0 | | x1 |   | 0 |
     | 0  1 | | x2 | = | 0 |
     

     From this, we get x₂ = 0, and x₁ is free. Let x₁ = s.

     The eigenvector is (s, 0) = s(1, 0).

   • For λ₂ = 3:

     (A - 3I)x = 0

     | -1  0 | | x1 |   | 0 |
     |  0  0 | | x2 | = | 0 |
     

     From this, we get x₁ = 0, and x₂ is free. Let x₂ = t.

     The eigenvector is (0, t) = t(0, 1).

3. Basis vectors:

   The basis for the eigenspace corresponding to λ₁ = 2 is {(1, 0)}. The basis for the eigenspace corresponding to λ₂ = 3 is {(0, 1)}.

Therefore, the bases for the eigenspaces are {(1, 0)} and {(0, 1)}.

Important Considerations

• Uniqueness: The basis of a subspace is not unique. There can be multiple sets of linearly independent vectors that span the same subspace.
• Dimension: The number of vectors in a basis for a subspace is called the dimension of the subspace. The dimension is unique for a given subspace.
• Zero Vector: The set containing only the zero vector, {0}, is a subspace, and its basis is the empty set {}. The dimension of this subspace is 0.
• Verification: Always verify that the set you find is indeed linearly independent and spans the subspace. This can be done by showing that any vector in the subspace can be written as a linear combination of the basis vectors and that no vector in the basis can be written as a linear combination of the others.

Conclusion

Finding the basis of a subspace is a critical skill in linear algebra, with applications in various fields such as engineering, physics, and computer science. This article has provided detailed methods and examples for finding the basis of a subspace when it is defined by a spanning set, equations, as the column space or null space of a matrix, or as an eigenspace. By understanding these methods and practicing with examples, you can confidently determine the basis of any subspace. Remember to always verify that your basis is linearly independent and spans the subspace to ensure accuracy.