Basis And Dimension Of Solution Space Homogeneous System Of Linear Equations

Jul 13, 2025 by qnaunigon 77 views

Find a Basis and Dimension of the Solution Space for a Homogeneous System of Linear Equations

Introduction

In linear algebra, understanding the solution space of a homogeneous system of linear equations is crucial. The solution space, also known as the null space or kernel, comprises all vectors that, when substituted into the system, result in the zero vector. Determining a basis for this solution space and its dimension provides insights into the system's structure and properties. This article will guide you through finding a basis and the dimension of the solution space for a given homogeneous system of linear equations. Let's consider the following system:

-x + y + z = 0
3x - y = 0
2x - 4y - 5z = 0

We aim to find (a) a basis for the solution space and (b) the dimension of the solution space. This involves transforming the system into a more manageable form, typically using Gaussian elimination or similar methods, and then identifying the free variables and expressing the solutions in terms of these variables. The resulting vectors will form the basis, and their count will give us the dimension.

Transforming the System into Row-Echelon Form

To find the solution space, our first step is to transform the given system of linear equations into row-echelon form. This form makes it easier to identify the leading and free variables, which are essential for determining the basis and dimension of the solution space. The system we are working with is:

-x + y + z = 0
3x - y = 0
2x - 4y - 5z = 0

We can represent this system as an augmented matrix:

[ -1  1  1 | 0 ]
[  3 -1  0 | 0 ]
[  2 -4 -5 | 0 ]

Now, we will perform row operations to bring the matrix into row-echelon form. Our goal is to create leading ones (1s) in each row and zeros below them. First, we can multiply the first row by -1 to get a leading 1:

[ 1 -1 -1 | 0 ]
[ 3 -1  0 | 0 ]
[ 2 -4 -5 | 0 ]

Next, we eliminate the 3 in the second row and the 2 in the third row by performing the following operations:

R2 = R2 - 3R1
R3 = R3 - 2R1

This gives us:

[  1 -1 -1 | 0 ]
[  0  2  3 | 0 ]
[  0 -2 -3 | 0 ]

Now, we can divide the second row by 2 to get a leading 1:

[  1 -1 -1 | 0 ]
[  0  1 3/2 | 0 ]
[  0 -2 -3 | 0 ]

Finally, we eliminate the -2 in the third row by performing the operation:

R3 = R3 + 2R2

This results in:

[  1 -1 -1 | 0 ]
[  0  1 3/2 | 0 ]
[  0  0  0 | 0 ]

The matrix is now in row-echelon form. This simplified form allows us to easily identify the relationships between the variables and find the solution space.

Identifying Free Variables and Expressing Solutions

After transforming the system into row-echelon form, the next crucial step is to identify the free variables and express the solutions in terms of these variables. The row-echelon form of our system is:

[  1 -1 -1 | 0 ]
[  0  1 3/2 | 0 ]
[  0  0  0 | 0 ]

This corresponds to the following system of equations:

x - y - z = 0
y + (3/2)z = 0

From this, we can see that z is a free variable because it does not correspond to a leading 1 in the row-echelon form. The variables x and y are basic variables because they do correspond to leading 1s. We will express x and y in terms of z.

From the second equation, we have:

y = -(3/2)z

Substituting this into the first equation, we get:

x - (-(3/2)z) - z = 0
x + (3/2)z - z = 0
x + (1/2)z = 0
x = -(1/2)z

Thus, the general solution can be written in terms of the free variable z as:

(x, y, z) = (-1/2 z, -3/2 z, z)

This representation allows us to express the solution space as a set of vectors, which is essential for finding a basis.

Finding a Basis for the Solution Space

Now that we have expressed the general solution in terms of the free variable z, we can proceed to find a basis for the solution space. The general solution is given by:

(x, y, z) = (-1/2 z, -3/2 z, z)

We can rewrite this as a linear combination of vectors by factoring out z:

(x, y, z) = z(-1/2, -3/2, 1)

This shows that the solution space is spanned by the vector (-1/2, -3/2, 1). To eliminate fractions, we can multiply this vector by 2, which gives us the vector (-1, -3, 2). This vector still spans the same space and is linearly independent since it is a single non-zero vector.

Therefore, a basis for the solution space is the set containing this single vector:

{(-1, -3, 2)}

A basis is a set of linearly independent vectors that span the solution space. In this case, the single vector (-1, -3, 2) forms the basis. This basis allows us to represent any solution in the solution space as a scalar multiple of this vector. Understanding the basis is crucial for characterizing the solution space and its properties.

Determining the Dimension of the Solution Space

Having found a basis for the solution space, we can now determine its dimension. The dimension of a vector space is the number of vectors in a basis for that space. In our case, the basis for the solution space is:

{(-1, -3, 2)}

This basis contains only one vector. Therefore, the dimension of the solution space is 1. This means that the solution space is a one-dimensional subspace of ℝ³, which is a line passing through the origin.

The dimension of the solution space is also equal to the number of free variables in the system. In our case, we had one free variable, z, which confirms that the dimension of the solution space is 1. Understanding the dimension provides crucial information about the structure of the solution space and its relationship to the original system of equations. A higher dimension indicates a larger solution space with more degrees of freedom.

Conclusion

In this article, we successfully found a basis and determined the dimension of the solution space for the given homogeneous system of linear equations:

-x + y + z = 0
3x - y = 0
2x - 4y - 5z = 0

We transformed the system into row-echelon form, identified the free variable z, and expressed the solutions in terms of z. This allowed us to find a basis for the solution space, which is the set {(-1, -3, 2)}. The dimension of the solution space is 1, indicating that the solutions form a line in three-dimensional space.

Understanding the basis and dimension of the solution space provides valuable insights into the nature of the system of equations. It allows us to characterize the set of all possible solutions and understand the relationships between the variables. This process is fundamental in linear algebra and has applications in various fields, including engineering, computer science, and economics. Mastering these concepts enables us to solve more complex problems and gain a deeper understanding of linear systems.

Key takeaways from this article include:

The process of transforming a system of linear equations into row-echelon form.
Identifying free variables and expressing solutions in terms of these variables.
Finding a basis for the solution space.
Determining the dimension of the solution space.

By following these steps, you can effectively analyze homogeneous systems of linear equations and understand their solution spaces.