Finding Basis And Dimension Of Solution Space For Ax=0

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Introduction

In linear algebra, finding the basis and dimension of the solution space (also known as the null space) of a matrix equation Ax = 0 is a fundamental problem. This article delves into a comprehensive approach to solving this problem, providing a step-by-step guide with a specific example. Understanding the null space and its properties is crucial for various applications, including solving systems of linear equations, understanding the structure of linear transformations, and more. We will explore the process of reducing a matrix to its reduced row-echelon form, identifying free variables, and constructing a basis for the null space. The dimension of the null space, also known as the nullity, provides important information about the solutions to the homogeneous equation Ax = 0.

Problem Statement

Given the matrix:

A=[25−3−2−2−32−513−22−1−535]A=\begin{bmatrix} 2 & 5 & -3 & -2 \\ -2 & -3 & 2 & -5 \\ 1 & 3 & -2 & 2 \\ -1 & -5 & 3 & 5 \end{bmatrix}

Our objective is to find a basis for the solution space (null space) of the equation Ax = 0 and determine the dimension of this space.

Methodology

The solution involves the following key steps:

  1. Reduce the matrix A to its reduced row-echelon form (RREF). This form simplifies the matrix while preserving the solution space.
  2. Identify the pivot columns and free variables. Pivot columns correspond to leading variables, while free variables can take any value.
  3. Express the leading variables in terms of the free variables. This step provides the general solution to Ax = 0.
  4. Construct the basis vectors for the null space. Each free variable corresponds to a basis vector, which is obtained by setting the free variable to 1 and the other free variables to 0.
  5. Determine the dimension of the null space. The dimension is equal to the number of basis vectors, which is the number of free variables.

Step-by-Step Solution

1. Reduce the Matrix to Reduced Row-Echelon Form (RREF)

We begin by applying Gaussian elimination to transform matrix A into its RREF. This involves a series of elementary row operations.

A=[25−3−2−2−32−513−22−1−535]A=\begin{bmatrix} 2 & 5 & -3 & -2 \\ -2 & -3 & 2 & -5 \\ 1 & 3 & -2 & 2 \\ -1 & -5 & 3 & 5 \end{bmatrix}

First, we swap Row 1 and Row 3 to get a smaller number in the first pivot position:

[13−22−2−32−525−3−2−1−535]\begin{bmatrix} 1 & 3 & -2 & 2 \\ -2 & -3 & 2 & -5 \\ 2 & 5 & -3 & -2 \\ -1 & -5 & 3 & 5 \end{bmatrix}

Next, we perform the following row operations:

  • R2←R2+2R1R_2 \leftarrow R_2 + 2R_1
  • R3←R3−2R1R_3 \leftarrow R_3 - 2R_1
  • R4←R4+R1R_4 \leftarrow R_4 + R_1

This yields:

[13−2203−2−10−11−60−217]\begin{bmatrix} 1 & 3 & -2 & 2 \\ 0 & 3 & -2 & -1 \\ 0 & -1 & 1 & -6 \\ 0 & -2 & 1 & 7 \end{bmatrix}

Now, we swap Row 2 and Row 3:

[13−220−11−603−2−10−217]\begin{bmatrix} 1 & 3 & -2 & 2 \\ 0 & -1 & 1 & -6 \\ 0 & 3 & -2 & -1 \\ 0 & -2 & 1 & 7 \end{bmatrix}

Multiply Row 2 by -1:

[13−2201−1603−2−10−217]\begin{bmatrix} 1 & 3 & -2 & 2 \\ 0 & 1 & -1 & 6 \\ 0 & 3 & -2 & -1 \\ 0 & -2 & 1 & 7 \end{bmatrix}

Perform the following row operations:

  • R1←R1−3R2R_1 \leftarrow R_1 - 3R_2
  • R3←R3−3R2R_3 \leftarrow R_3 - 3R_2
  • R4←R4+2R2R_4 \leftarrow R_4 + 2R_2

This gives us:

[101−1601−16001−1900−119]\begin{bmatrix} 1 & 0 & 1 & -16 \\ 0 & 1 & -1 & 6 \\ 0 & 0 & 1 & -19 \\ 0 & 0 & -1 & 19 \end{bmatrix}

Next, we add Row 3 to Row 4:

[101−1601−16001−190000]\begin{bmatrix} 1 & 0 & 1 & -16 \\ 0 & 1 & -1 & 6 \\ 0 & 0 & 1 & -19 \\ 0 & 0 & 0 & 0 \end{bmatrix}

Finally, perform the following row operations:

  • R1←R1−R3R_1 \leftarrow R_1 - R_3
  • R2←R2+R3R_2 \leftarrow R_2 + R_3

We obtain the reduced row-echelon form:

[1003010−13001−190000]\begin{bmatrix} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & -13 \\ 0 & 0 & 1 & -19 \\ 0 & 0 & 0 & 0 \end{bmatrix}

2. Identify Pivot Columns and Free Variables

In the RREF, the pivot columns are the first, second, and third columns (corresponding to variables x1x_1, x2x_2, and x3x_3). The fourth column does not have a pivot, so x4x_4 is a free variable.

3. Express Leading Variables in Terms of Free Variables

From the RREF, we can write the following equations:

  • x1+3x4=0  ⟹  x1=−3x4x_1 + 3x_4 = 0 \implies x_1 = -3x_4
  • x2−13x4=0  ⟹  x2=13x4x_2 - 13x_4 = 0 \implies x_2 = 13x_4
  • x3−19x4=0  ⟹  x3=19x4x_3 - 19x_4 = 0 \implies x_3 = 19x_4

So the general solution is:

[x1x2x3x4]=[−3x413x419x4x4]=x4[−313191]\begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} -3x_4 \\ 13x_4 \\ 19x_4 \\ x_4 \end{bmatrix} = x_4 \begin{bmatrix} -3 \\ 13 \\ 19 \\ 1 \end{bmatrix}

4. Construct Basis Vectors for the Null Space

Since we have one free variable (x4x_4), there will be one basis vector. Setting x4=1x_4 = 1, we get the basis vector:

[−313191]\begin{bmatrix} -3 \\ 13 \\ 19 \\ 1 \end{bmatrix}

Thus, the basis for the null space is:

{[−313191]}\left\{ \begin{bmatrix} -3 \\ 13 \\ 19 \\ 1 \end{bmatrix} \right\}

5. Determine the Dimension of the Null Space

The dimension of the null space is the number of basis vectors, which is 1 in this case. This is because there is one free variable.

Conclusion

In summary, for the given matrix A, the basis for the solution space (null space) of Ax = 0 is:

{[−313191]}\left\{ \begin{bmatrix} -3 \\ 13 \\ 19 \\ 1 \end{bmatrix} \right\}

The dimension of the null space (nullity) is 1. This means that there is one linearly independent solution to the homogeneous equation Ax = 0. Understanding these concepts is crucial for solving linear systems and comprehending the properties of linear transformations. By following the steps outlined, one can systematically determine the basis and dimension of the solution space for any given matrix.

This detailed solution not only provides the answer but also illuminates the process, making it easier to apply the same methodology to other problems in linear algebra. The ability to find the basis and dimension of the null space is a powerful tool in mathematical analysis and its various applications.