Finding The Basis For Row Space And Rank Of A Matrix

by ADMIN 53 views

Introduction

In the realm of linear algebra, understanding the properties of matrices is crucial. Among these properties, the row space and the rank of a matrix hold significant importance. The row space provides insights into the linear combinations of the rows, while the rank indicates the number of linearly independent rows. In this article, we will delve into the process of finding a basis for the row space and determining the rank of a given matrix. Our focus will be on a specific example matrix, but the methods discussed are applicable to matrices in general. Understanding these concepts is fundamental for solving systems of linear equations, performing matrix decompositions, and grasping the structure of vector spaces.

The process of finding a basis for the row space and the rank of a matrix involves performing row operations to transform the matrix into its row-echelon form or reduced row-echelon form. The non-zero rows in the row-echelon form then form a basis for the row space. The number of these non-zero rows is the rank of the matrix. This method is efficient and provides a clear understanding of the matrix's structure. We will walk through the steps, explaining the logic behind each operation and how it contributes to the final result. This comprehensive approach aims to solidify your understanding of these concepts and their applications in various mathematical and computational contexts.

By the end of this discussion, you will not only be able to compute the row space and rank of a matrix but also appreciate the significance of these concepts in broader applications. Whether you are a student learning linear algebra or a professional applying these techniques, this article will provide you with a solid foundation. We will also touch upon the connection between the row space, column space, and null space, further enriching your understanding of matrix properties. Let's embark on this journey to unravel the intricacies of matrix analysis and its practical implications.

Problem Statement

Consider the following matrix:

[−2−44936−6−4−2−445]\begin{bmatrix} -2 & -4 & 4 & 9 \\ 3 & 6 & -6 & -4 \\ -2 & -4 & 4 & 5 \end{bmatrix}

Our goal is to find:

(a) A basis for the row space of the matrix.

(b) The rank of the matrix.

Solution: Finding a Basis for the Row Space

The row space of a matrix is the vector space spanned by its row vectors. To find a basis for the row space, we need to identify a set of linearly independent vectors that span the same space as the rows of the original matrix. This is typically done by performing row operations to bring the matrix into its row-echelon form or reduced row-echelon form. The non-zero rows in the row-echelon form will then form a basis for the row space. Let's proceed step by step.

Step 1: Transform the Matrix to Row-Echelon Form

Our first step involves using elementary row operations to transform the given matrix into row-echelon form. These operations include:

  1. Swapping two rows.
  2. Multiplying a row by a non-zero scalar.
  3. Adding a multiple of one row to another row.

These operations do not change the row space of the matrix, which is crucial for our goal of finding a basis.

Let's start by focusing on the first column. We want to eliminate the entries below the first entry. First, we can multiply the first row by -1/2 to simplify the calculations:

R1→−12R1R_1 \rightarrow -\frac{1}{2}R_1

This gives us:

[12−2−9/236−6−4−2−445]\begin{bmatrix} 1 & 2 & -2 & -9/2 \\ 3 & 6 & -6 & -4 \\ -2 & -4 & 4 & 5 \end{bmatrix}

Next, we eliminate the 3 in the second row by replacing the second row with R2−3R1R_2 - 3R_1:

R2→R2−3R1R_2 \rightarrow R_2 - 3R_1

[12−2−9/200023/2−2−445]\begin{bmatrix} 1 & 2 & -2 & -9/2 \\ 0 & 0 & 0 & 23/2 \\ -2 & -4 & 4 & 5 \end{bmatrix}

Now, we eliminate the -2 in the third row by replacing the third row with R3+2R1R_3 + 2R_1:

R3→R3+2R1R_3 \rightarrow R_3 + 2R_1

[12−2−9/200023/2000−4]\begin{bmatrix} 1 & 2 & -2 & -9/2 \\ 0 & 0 & 0 & 23/2 \\ 0 & 0 & 0 & -4 \end{bmatrix}

Step 2: Further Row Operations

Now, we want to get a leading 1 in the second row. We can multiply the second row by 2/23:

R2→223R2R_2 \rightarrow \frac{2}{23}R_2

[12−2−9/20001000−4]\begin{bmatrix} 1 & 2 & -2 & -9/2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & -4 \end{bmatrix}

Next, we eliminate the -4 in the third row by replacing it with R3+4R2R_3 + 4R_2:

R3→R3+4R2R_3 \rightarrow R_3 + 4R_2

[12−2−9/200010000]\begin{bmatrix} 1 & 2 & -2 & -9/2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}

Step 3: Identify the Basis Vectors

The matrix is now in row-echelon form. The non-zero rows are:

[12−2−9/2]\begin{bmatrix} 1 & 2 & -2 & -9/2 \end{bmatrix}

and

[0001]\begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix}

Thus, a basis for the row space is:

{[12−2−9/2],[0001]}\left\{\begin{bmatrix} 1 & 2 & -2 & -9/2 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix}\right\}

This set of vectors is linearly independent and spans the same row space as the original matrix. The basis for the row space is a fundamental concept in linear algebra, providing a concise representation of the space spanned by the rows of the matrix.

Solution: Determining the Rank of the Matrix

The rank of a matrix is the number of linearly independent rows (or columns) in the matrix. It is equal to the number of non-zero rows in the row-echelon form (or reduced row-echelon form) of the matrix. In our case, the row-echelon form of the matrix is:

[12−2−9/200010000]\begin{bmatrix} 1 & 2 & -2 & -9/2 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{bmatrix}

There are two non-zero rows. Therefore, the rank of the matrix is 2. The rank provides critical information about the matrix, including the dimensionality of the row and column spaces.

(a) Basis for the Row Space

Based on our row reduction, a basis for the row space is:

{[12−2−9/2],[0001]}\left\{\begin{bmatrix} 1 & 2 & -2 & -9/2 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 0 & 1 \end{bmatrix}\right\}

This basis consists of two linearly independent vectors that span the row space of the original matrix. The process of finding a basis for the row space often involves transforming the matrix to its row-echelon form, which simplifies the identification of linearly independent rows.

(b) The Rank of the Matrix

The rank of the matrix is the number of non-zero rows in its row-echelon form. In our case, there are two non-zero rows. Therefore, the rank of the matrix is 2. Understanding the rank of a matrix is essential, as it provides insights into the matrix's invertibility and the solutions of linear systems.

Conclusion

In this article, we successfully found a basis for the row space and determined the rank of the given matrix. We started by transforming the matrix into its row-echelon form using elementary row operations. The non-zero rows in the row-echelon form then provided us with a basis for the row space. The number of these non-zero rows gave us the rank of the matrix. These concepts are fundamental in linear algebra and have wide-ranging applications in various fields, including engineering, computer science, and economics. The rank of the matrix and the basis for its row space are key indicators of the matrix's properties and behavior. Understanding these concepts is vital for anyone working with matrices and linear systems.

The ability to find a basis for the row space and determine the rank is crucial for solving linear systems, understanding matrix transformations, and analyzing vector spaces. The row space represents the span of the rows of the matrix, while the rank indicates the number of linearly independent rows. These concepts are interconnected and provide a comprehensive view of the matrix's structure. Further exploration of these topics will reveal even more applications and connections within linear algebra and beyond. The row space and rank are not just theoretical constructs; they are powerful tools that enable us to solve real-world problems involving linear relationships. Whether you are designing algorithms, analyzing data, or modeling physical systems, a solid grasp of these concepts will prove invaluable. This article has provided a detailed walkthrough of the process, but the journey of understanding linear algebra is an ongoing one. Continue to practice, explore different examples, and challenge yourself with more complex problems. The more you engage with these concepts, the deeper your understanding will become.