Finding Basis And Dimension Of Subspaces In P3(t) And Matrix Spaces

Introduction

In linear algebra, understanding the basis and dimension of a subspace is crucial for grasping the structure and properties of vector spaces. This article delves into the process of finding the basis and dimension of subspaces, focusing on polynomial spaces P₃(t) and matrix spaces. We will explore how to determine the linearly independent vectors that span a given subspace and how to count these vectors to find the dimension. This exploration will involve techniques such as forming matrices from the coefficients of polynomials or entries of matrices and then performing row reduction to identify pivot columns, which correspond to the basis vectors. Through detailed examples, we aim to provide a comprehensive understanding of these concepts and their applications.

1. Find the Basis and Dimension of the Subspace W

In this section, we will explore how to find the basis and dimension of a subspace W spanned by a set of vectors. We will consider examples in the polynomial space P₃(t) and in matrix spaces. The key idea is to determine a set of linearly independent vectors that span the subspace, which forms the basis. The number of these vectors then gives the dimension of the subspace. This involves forming a matrix with the coefficients of the vectors and reducing it to row-echelon form to identify pivot columns, which correspond to the basis vectors. Let's dive into the specific examples.

a. Subspace W Spanned by Polynomials u, v, and w in P₃(t)

Let's consider the subspace W in P₃(t) spanned by the polynomials:

• u = t³ + 2t² - 2t + 1
• v = t³ + 3t² - 3t + 4
• w = 2t³ + t² - 7t - 7

To find a basis for W and its dimension, we first represent these polynomials as vectors using their coefficients. The coefficients of the polynomials u, v, and w can be written as the column vectors (1, 2, -2, 1), (1, 3, -3, 4), and (2, 1, -7, -7), respectively. We then form a matrix A with these vectors as columns:

A = 
| 1  1  2 |
| 2  3  1 |
|-2 -3 -7 |
| 1  4 -7 |

Now, we perform row reduction to find the pivot columns, which correspond to the basis vectors. Applying row operations, we get:

1. Subtract 2 times the first row from the second row, add 2 times the first row to the third row, and subtract the first row from the fourth row:

A ~ 
| 1  1  2 |
| 0  1 -3 |
| 0 -1 -3 |
| 0  3 -9 |

2. Add the second row to the third row and subtract 3 times the second row from the fourth row:

A ~ 
| 1  1  2 |
| 0  1 -3 |
| 0  0 -6 |
| 0  0  0 |

The matrix is now in row-echelon form. The pivot columns are the first, second, and third columns. These columns correspond to the polynomials u, v, and w. Since there are three pivot columns, the basis for the subspace W consists of the polynomials corresponding to these columns. Therefore, the basis for W is {u, v, w}, and the dimension of W is 3. This means that any polynomial in W can be written as a linear combination of u, v, and w.

In summary, finding the basis and dimension involves forming a matrix from the polynomial coefficients, reducing it to row-echelon form, and identifying pivot columns. The pivot columns indicate the linearly independent polynomials that form the basis, and the number of these polynomials is the dimension of the subspace. This process is fundamental in understanding the structure of subspaces in linear algebra.

b. Subspace W Spanned by Polynomials u, v, and w in P₃(t)

Consider the subspace W in P₃(t) spanned by the polynomials:

• u = t³ + t² - 3t + 2
• v = 2t³ + t² + t - 4
• w = 4t³ + 3t² - 5t + 2

To find a basis for W and its dimension, we follow a similar procedure as in the previous example. We represent these polynomials as vectors using their coefficients. The coefficients of the polynomials u, v, and w can be written as the column vectors (1, 1, -3, 2), (2, 1, 1, -4), and (4, 3, -5, 2), respectively. We then form a matrix A with these vectors as columns:

A = 
| 1  2  4 |
| 1  1  3 |
|-3  1 -5 |
| 2 -4  2 |

We perform row reduction to find the pivot columns. Applying row operations, we get:

1. Subtract the first row from the second row, add 3 times the first row to the third row, and subtract 2 times the first row from the fourth row:

A ~ 
| 1  2  4 |
| 0 -1 -1 |
| 0  7  7 |
| 0 -8 -6 |

2. Multiply the second row by -1, add 7 times the second row to the third row, and subtract 8 times the second row from the fourth row:

A ~ 
| 1  2  4 |
| 0  1  1 |
| 0  0  0 |
| 0  0  2 |

3. Swap the third and fourth rows:

A ~ 
| 1  2  4 |
| 0  1  1 |
| 0  0  2 |
| 0  0  0 |

The matrix is now in row-echelon form. The pivot columns are the first, second, and third columns. These columns correspond to the polynomials u, v, and w. Since there are three pivot columns, the basis for the subspace W consists of the polynomials corresponding to these columns. Therefore, a basis for W is {u, v, w}, and the dimension of W is 3. This indicates that any polynomial in W can be expressed as a linear combination of u, v, and w.

In essence, determining the basis and dimension involves creating a matrix from the polynomial coefficients, reducing it to row-echelon form, and identifying the pivot columns. The pivot columns represent the linearly independent polynomials forming the basis, while the number of these polynomials defines the dimension of the subspace. This process is crucial for understanding the composition of subspaces in linear algebra, ensuring that we can express any element within the subspace using the identified basis.

c. Subspace W Spanned by Matrices A, B, and C

Consider the subspace W spanned by the following matrices:

• A = | 1 -5 | |-4 2 |

• B = |-2 1 | | 3 -1 |

• C = |-6 -3 | | 5 -3 |

To find a basis for W and its dimension, we represent these matrices as column vectors. We can represent each matrix as a 4-dimensional vector by stacking its columns. This gives us the vectors (1, -4, -5, 2), (-2, 3, 1, -1), and (-6, 5, -3, -3) corresponding to matrices A, B, and C, respectively. We form a matrix with these vectors as columns:

Matrix = 
| 1 -2 -6 |
|-4  3  5 |
|-5  1 -3 |
| 2 -1 -3 |

Now, we perform row reduction to find the pivot columns.

1. Add 4 times the first row to the second row, add 5 times the first row to the third row, and subtract 2 times the first row from the fourth row:

Matrix ~ 
| 1 -2 -6 |
| 0 -5 -19|
| 0 -9 -33|
| 0  3  9 |

2. Divide the second row by -5:

Matrix ~ 
| 1 -2 -6 |
| 0  1  19/5|
| 0 -9 -33|
| 0  3  9 |

3. Add 9 times the second row to the third row, and subtract 3 times the second row from the fourth row:

Matrix ~ 
| 1 -2 -6 |
| 0  1  19/5|
| 0  0 -6/5 |
| 0  0 -12/5|

4. Multiply the third row by -5/6:

Matrix ~ 
| 1 -2 -6 |
| 0  1  19/5|
| 0  0  1 |
| 0  0 -12/5|

5. Add 12/5 times the third row to the fourth row:

Matrix ~ 
| 1 -2 -6 |
| 0  1  19/5|
| 0  0  1 |
| 0  0  0 |

The matrix is now in row-echelon form. The pivot columns are the first, second, and third columns. These correspond to the matrices A, B, and C. Thus, a basis for the subspace W is {A, B, C}, and the dimension of W is 3. This indicates that any matrix in W can be expressed as a linear combination of A, B, and C.

In summary, to determine the basis and dimension for a subspace spanned by matrices, we convert each matrix into a column vector and form a matrix with these vectors. Reducing this matrix to row-echelon form and identifying pivot columns provides the linearly independent matrices that form the basis. The count of these matrices gives the dimension of the subspace. This method allows us to understand how matrices combine within a vector space, which is a fundamental concept in linear algebra.

Conclusion

Finding the basis and dimension of subspaces is a fundamental concept in linear algebra. This article has demonstrated how to determine these properties for subspaces spanned by polynomials in P₃(t) and matrices. The method involves representing the polynomials or matrices as vectors, forming a matrix with these vectors, and then reducing the matrix to row-echelon form. The pivot columns in the row-echelon form correspond to the linearly independent vectors that form the basis of the subspace. The number of vectors in the basis gives the dimension of the subspace. Understanding these concepts is essential for solving various problems in linear algebra, including linear transformations, eigenvalue problems, and more. By mastering these techniques, one can gain a deeper insight into the structure and properties of vector spaces and subspaces.

By following the detailed steps and examples provided in this article, readers should now be well-equipped to find the basis and dimension of various subspaces. These skills are crucial for further studies in mathematics, physics, engineering, and computer science, where linear algebra plays a vital role. The ability to identify a basis and determine the dimension of a subspace allows for a more streamlined and efficient approach to problem-solving and a greater understanding of the underlying mathematical structures.