Finding The Basis Of A Subspace

Delving into the realm of linear algebra, finding the basis of a subspace is a fundamental skill, allowing us to understand the structure and dimensionality of vector spaces. A basis provides the most efficient way to represent every vector within that subspace, using a minimal set of linearly independent vectors. This article provides a comprehensive guide on how to find the basis of a subspace, covering various scenarios and providing practical examples to solidify understanding.

What is a Subspace, Basis, and Linear Independence?

Before diving into the process, let's clarify some key concepts:

• Subspace: A subspace is a subset of a vector space that itself satisfies the properties of a vector space. Specifically, it must be closed under vector addition and scalar multiplication.
• Basis: A basis of a subspace is a set of linearly independent vectors that span the entire subspace. "Span" means that any vector in the subspace can be written as a linear combination of the basis vectors.
• 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. Equivalently, the only solution to the equation c1v1 + c2v2 + ... + cnvn = 0 (where c1, c2, ..., cn are scalars and v1, v2, ..., vn are the vectors) is c1 = c2 = ... = cn = 0.

In simpler terms, a basis is the smallest possible set of vectors you need to describe every vector in a subspace without redundancy.

Steps to Find the Basis of a Subspace

The general process for finding the basis of a subspace depends on how the subspace is defined. Here are common scenarios and the corresponding steps:

1. Subspace Defined by a Spanning Set

Often, a subspace V is defined as the span of a set of vectors, say {v1, v2, ..., vn}. This means V = span{v1, v2, ..., vn}. To find a basis for V:

Step 1: Form a Matrix

Construct a matrix A whose columns are the vectors v1, v2, ..., vn.

A = [ v1  v2  ...  vn ]

Step 2: Row Reduce the Matrix

Use Gaussian elimination (row reduction) to transform the matrix A into its reduced row echelon form (RREF). The RREF is a unique form that helps identify linearly independent columns.

Step 3: Identify Pivot Columns

Identify the pivot columns in the RREF. Pivot columns are the columns that contain a leading 1 (the first non-zero entry in a row).

Step 4: Extract Corresponding Vectors

The vectors in the original matrix A that correspond to the pivot columns in the RREF form the basis of the subspace V.

Example:

Suppose V = span{v1, v2, v3, v4} where:

• v1 = [1, 2, 3]
• v2 = [2, 4, 6]
• v3 = [1, 0, 1]
• v4 = [0, 2, 2]

Step 1: Form the Matrix

A = [ 1  2  1  0 ]
    [ 2  4  0  2 ]
    [ 3  6  1  2 ]

Step 2: Row Reduce to RREF

After performing row reduction, we might get:

RREF(A) = [ 1  2  0  1 ]
          [ 0  0  1  -1 ]
          [ 0  0  0  0 ]

Step 3: Identify Pivot Columns

The pivot columns are the first and third columns (containing the leading 1's).

Step 4: Extract Corresponding Vectors

The basis of V is {v1, v3} or {[1, 2, 3], [1, 0, 1]}. This means that any vector in V can be expressed as a linear combination of v1 and v3. Notice that v2 and v4 are redundant because they are linear combinations of v1 and v3 (in fact, v2 = 2v1 and v4 = v1 - v3).

2. Subspace Defined by Equations

Sometimes, a subspace is defined as the set of all vectors that satisfy a set of linear equations. For instance, V = { [x, y, z] ∈ R³ | x + y + z = 0 and x - y = 0 }. To find a basis for such a subspace:

Step 1: Solve the System of Equations

Solve the system of linear equations to express the variables in terms of free variables.

Step 2: Express General Solution as a Linear Combination

Write the general solution as a linear combination of vectors, where the coefficients are the free variables.

Step 3: The Basis Vectors

The vectors that multiply the free variables in the linear combination form the basis of the subspace.

Example:

V = { [x, y, z] ∈ R³ | x + y + z = 0 and x - y = 0 }

Step 1: Solve the System of Equations

From x - y = 0, we get x = y. Substituting into the first equation:

y + y + z = 0 => 2y + z = 0 => z = -2y

So, x = y and z = -2y.

Step 2: Express General Solution

The general solution can be written as:

[x, y, z] = [y, y, -2y] = y[1, 1, -2]

Step 3: The Basis Vectors

The basis of V is {[1, 1, -2]}. The subspace is a line through the origin in R³ defined by the direction vector [1, 1, -2].

3. Null Space (Kernel) of a Matrix

The null space (or kernel) of a matrix A is the set of all vectors x such that Ax = 0. In other words, null(A) = { x | Ax = 0 }. The null space is a subspace. To find a basis for the null space:

Step 1: Solve the Homogeneous Equation Ax = 0

Solve the homogeneous system of linear equations represented by Ax = 0. This usually involves row reducing the augmented matrix [A | 0] to RREF.

Step 2: Express General Solution as a Linear Combination

Write the general solution in terms of free variables.

Step 3: The Basis Vectors

The vectors that multiply the free variables in the linear combination form the basis of the null space.

Example:

Let A = [ 1 2 3 ] [ 2 4 6 ]

Find a basis for null(A).

Step 1: Solve Ax = 0

We want to solve:

[ 1 2 3 ] [x] = [0] [ 2 4 6 ] [y] = [0] [z] [0]

Row reducing the augmented matrix [A | 0]:

[ 1  2  3 | 0 ]
[ 2  4  6 | 0 ]  -->  [ 1  2  3 | 0 ]
                      [ 0  0  0 | 0 ]

From the RREF, we have the equation x + 2y + 3z = 0, so x = -2y - 3z.

Step 2: Express General Solution

The general solution is:

[x, y, z] = [-2y - 3z, y, z] = y[-2, 1, 0] + z[-3, 0, 1]

Step 3: The Basis Vectors

The basis of null(A) is {[-2, 1, 0], [-3, 0, 1]}. The null space is a plane in R³ passing through the origin.

4. Column Space (Range) of a Matrix

The column space (or range) of a matrix A is the span of its column vectors. In other words, col(A) = { Ax | x ∈ Rⁿ }, where A is an m x n matrix. The column space is a subspace. To find a basis for the column space:

Step 1: Row Reduce the Matrix

Row reduce the matrix A to its RREF.

Step 2: Identify Pivot Columns

Identify the pivot columns in the RREF.

Step 3: Extract Corresponding Vectors from the Original Matrix

The vectors in the original matrix A that correspond to the pivot columns in the RREF form the basis of the column space. This is the same procedure as finding the basis of a subspace defined by a spanning set (described in Section 1).

Example:

Let A = [ 1 2 1 ] [ 2 4 0 ] [ 3 6 1 ]

Find a basis for col(A).

Step 1: Row Reduce the Matrix

Row reducing A to RREF gives (as seen in the first example):

RREF(A) = [ 1  2  0 ]
          [ 0  0  1 ]
          [ 0  0  0 ]

Step 2: Identify Pivot Columns

The pivot columns are the first and third columns.

Step 3: Extract Corresponding Vectors from the Original Matrix

The basis of col(A) is {[1, 2, 3], [1, 0, 1]}.

Important Considerations and Common Mistakes

• RREF is Crucial: Using the reduced row echelon form (RREF) is essential for accurately identifying pivot columns and linear independence. Incorrect row reduction will lead to an incorrect basis.
• Pivot Columns from Original Matrix: Remember to extract the basis vectors from the original matrix, not the RREF. The RREF only helps identify which columns of the original matrix are linearly independent.
• Uniqueness: The basis of a subspace is not unique. There can be multiple sets of linearly independent vectors that span the same subspace. However, the number of vectors in any basis of a given subspace is always the same. This number is called the dimension of the subspace.
• Dimension: The dimension of a subspace is the number of vectors in its basis. This is a fundamental property of the subspace and is independent of the choice of basis.
• Zero Vector: The zero vector is never part of a basis. A basis consists only of linearly independent vectors. If you include the zero vector, the set becomes linearly dependent.
• Checking Linear Independence: After finding a potential basis, it's always a good idea to check that the vectors are indeed linearly independent. You can do this by setting up the equation c1v1 + c2v2 + ... + cnvn = 0 and verifying that the only solution is c1 = c2 = ... = cn = 0.
• Checking Spanning: You can verify that the basis vectors span the subspace by showing that any vector in the subspace can be written as a linear combination of the basis vectors. This is often easier if you already know the definition of the subspace (e.g., defined by equations).

Examples in Different Vector Spaces

The principles described above apply to various vector spaces, not just Rⁿ. Here are a couple of examples:

Example 1: Polynomials

Let V be the subspace of P₂ (the vector space of polynomials of degree at most 2) spanned by the polynomials:

• p1(t) = 1 + t + t²
• p2(t) = 2 + 2t + 2t²
• p3(t) = 1 + t

Find a basis for V.

We can represent each polynomial as a vector of its coefficients:

• p1(t) <-> [1, 1, 1]
• p2(t) <-> [2, 2, 2]
• p3(t) <-> [1, 1, 0]

Now, form a matrix with these vectors as columns:

A = [ 1  2  1 ]
    [ 1  2  1 ]
    [ 1  2  0 ]

Row reducing to RREF:

RREF(A) = [ 1  2  0 ]
          [ 0  0  1 ]
          [ 0  0  0 ]

The pivot columns are the first and third. Therefore, the basis for V is {1 + t + t², 1 + t} (the original polynomials corresponding to the first and third columns).

Example 2: Matrices

Let V be the subspace of M₂₂(R) (the vector space of 2x2 matrices with real entries) spanned by the matrices:

• M1 = [ 1 0 ] [ 0 0 ]

• M2 = [ 0 1 ] [ 0 0 ]

• M3 = [ 0 0 ] [ 1 0 ]

• M4 = [ 0 0 ] [ 0 1 ]

This is, in fact, all of M₂₂(R). Show that {M1, M2, M3, M4} forms a basis for M₂₂(R).

To demonstrate this, we need to show that these matrices are linearly independent and span M₂₂(R).

• Linear Independence: Suppose c1M1 + c2M2 + c3M3 + c4M4 = 0 (where 0 is the zero matrix). This means:

[ c1  c2 ] = [ 0  0 ]
[ c3  c4 ]   [ 0  0 ]

This implies c1 = c2 = c3 = c4 = 0. Thus, the matrices are linearly independent.

• Spanning: Let A be any 2x2 matrix:

A = [ a  b ]
    [ c  d ]

We can write A as a linear combination of M1, M2, M3, M4:

A = aM1 + bM2 + cM3 + dM4

Therefore, {M1, M2, M3, M4} spans M₂₂(R).

Since they are linearly independent and span M₂₂(R), {M1, M2, M3, M4} forms a basis for M₂₂(R). This is often called the "standard basis" for M₂₂(R).

Advanced Techniques and Applications

While the techniques described above cover most common scenarios, some situations require more advanced approaches:

• Gram-Schmidt Process: If you need to find an orthogonal basis (where the basis vectors are mutually orthogonal), you can use the Gram-Schmidt process to transform any basis into an orthogonal basis.
• Eigenbases: In the context of linear transformations and eigenvalues, finding an eigenbasis (a basis consisting of eigenvectors) can simplify the analysis of the transformation.
• Applications in Data Science and Machine Learning: Basis vectors are used extensively in data science. For example, principal component analysis (PCA) relies on finding a new basis for your data by finding the eigenvectors of the covariance matrix. Representing your data in this new basis can reduce the dimensionality of your data while preserving most of the variance.

Conclusion

Finding the basis of a subspace is a cornerstone of linear algebra. By understanding the concepts of spanning sets, linear independence, and utilizing techniques like row reduction, you can effectively determine a minimal set of vectors that completely describe any vector within the subspace. Mastering these techniques provides a solid foundation for further exploration of linear algebra and its applications in various fields. Remember to practice with diverse examples to solidify your understanding and develop intuition for these fundamental concepts.