Finding Basis And Dimension Of Solution Space For Homogeneous Linear Equations
In linear algebra, understanding the solution space of a homogeneous system of linear equations is a fundamental concept. The solution space represents the set of all vectors that satisfy the given system. This article will guide you through the process of finding a basis for the solution space and determining its dimension, using the following homogeneous system as an example:
-x + y + z = 0
3x - y = 0
2x - 4y - 5z = 0
We will first explore the theoretical background, then move on to the step-by-step solution, and finally, discuss the significance of these concepts.
Understanding Homogeneous Systems and Solution Spaces
A homogeneous system of linear equations is a system where all the constant terms are zero. This means the system can be written in the form Ax = 0, where A is the coefficient matrix, x is the vector of variables, and 0 is the zero vector. Homogeneous systems always have at least one solution, the trivial solution, where all variables are zero. However, they may also have infinitely many non-trivial solutions, which form the solution space.
The solution space of a homogeneous system is a subspace of the vector space in which the solutions lie. This means it is closed under addition and scalar multiplication. A basis for the solution space is a set of linearly independent vectors that span the space. The dimension of the solution space is the number of vectors in the basis.
Steps to Find the Basis and Dimension
To find the basis and dimension of the solution space, we follow these steps:
- Write the augmented matrix: Represent the system of equations as an augmented matrix.
- Reduce to row-echelon form: Use Gaussian elimination or Gauss-Jordan elimination to transform the matrix into row-echelon form or reduced row-echelon form.
- Identify free variables: Determine the variables that correspond to columns without leading ones (pivots). These are the free variables.
- Express leading variables in terms of free variables: Write the leading variables (corresponding to columns with leading ones) in terms of the free variables.
- Form the general solution: Express the solution vector in terms of the free variables.
- Construct the basis: Decompose the general solution into a linear combination of vectors, where each vector corresponds to a free variable. These vectors form a basis for the solution space.
- Determine the dimension: The dimension of the solution space is equal to the number of vectors in the basis, which is the number of free variables.
Solving the Given System
Now, let's apply these steps to the given system:
-x + y + z = 0
3x - y = 0
2x - 4y - 5z = 0
1. Write the Augmented Matrix
The augmented matrix for the system is:
[ -1 1 1 | 0 ]
[ 3 -1 0 | 0 ]
[ 2 -4 -5 | 0 ]
2. Reduce to Row-Echelon Form
We will use Gaussian elimination to reduce the matrix to row-echelon form. First, we 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:
- R2 = R2 - 3R1
- R3 = R3 - 2R1
[ 1 -1 -1 | 0 ]
[ 0 2 3 | 0 ]
[ 0 -2 -3 | 0 ]
Now, we can see that the third row is the negative of the second row. We can eliminate the third row by adding the second row to it:
- R3 = R3 + R2
[ 1 -1 -1 | 0 ]
[ 0 2 3 | 0 ]
[ 0 0 0 | 0 ]
Finally, we divide the second row by 2 to get a leading 1:
- R2 = (1/2)R2
[ 1 -1 -1 | 0 ]
[ 0 1 3/2 | 0 ]
[ 0 0 0 | 0 ]
This is the row-echelon form of the matrix. We can further reduce it to reduced row-echelon form by eliminating the -1 in the first row:
- R1 = R1 + R2
[ 1 0 1/2 | 0 ]
[ 0 1 3/2 | 0 ]
[ 0 0 0 | 0 ]
This is the reduced row-echelon form.
3. Identify Free Variables
From the reduced row-echelon form, we can see that the leading variables are x and y, corresponding to the first and second columns. The free variable is z, corresponding to the third column.
4. Express Leading Variables in Terms of Free Variables
From the reduced row-echelon form, we can write the equations:
x + (1/2)z = 0
y + (3/2)z = 0
Solving for x and y in terms of z, we get:
x = -(1/2)z
y = -(3/2)z
5. Form the General Solution
The general solution vector can be written as:
[ x ] [ -(1/2)z ]
[ y ] = [ -(3/2)z ]
[ z ] [ z ]
We can rewrite this as:
[ x ] [ -1/2 ]
[ y ] = z [ -3/2 ]
[ z ] [ 1 ]
6. Construct the Basis
The vector that multiplies z forms a basis for the solution space:
v = [ -1/2 ]
[ -3/2 ]
[ 1 ]
We can multiply this vector by 2 to eliminate the fractions and get a simpler basis vector:
v = [ -1 ]
[ -3 ]
[ 2 ]
Thus, a basis for the solution space is {[-1, -3, 2]}.
7. Determine the Dimension
The dimension of the solution space is the number of vectors in the basis, which is 1 in this case. This corresponds to the number of free variables (z).
Results
(a) A basis for the solution space is {[-1, -3, 2]}.
(b) The dimension of the solution space is 1.
Conclusion
Finding the basis and dimension of the solution space of a homogeneous system of linear equations is a crucial skill in linear algebra. By following the steps outlined in this article, you can systematically determine the basis vectors and the dimension of the space. This understanding is essential for solving more complex problems in linear algebra and its applications in various fields, such as engineering, physics, and computer science. The solution space provides insights into the nature of the solutions and the structure of the system itself. Mastering these concepts will significantly enhance your ability to work with linear systems and vector spaces.
Importance of Understanding Solution Spaces
The concept of solution spaces extends beyond simple problem-solving; it provides a deeper understanding of the behavior of linear systems. The dimension of the solution space, also known as the nullity of the matrix, is a key characteristic that reveals the degrees of freedom within the system. A higher dimension indicates a larger set of possible solutions, reflecting a greater flexibility in the system's variables.
In practical applications, solution spaces are instrumental in fields like data analysis and machine learning. For instance, in dimensionality reduction techniques like Principal Component Analysis (PCA), understanding the solution space helps identify the most significant components of the data. This leads to more efficient data representation and improved model performance. Furthermore, in engineering, analyzing the solution space of a system of equations can help in designing stable and robust systems, ensuring that the system behaves predictably under various conditions.
Further Exploration
To deepen your understanding of homogeneous systems and solution spaces, consider exploring related topics such as:
- Linear Independence and Span: These concepts are fundamental to understanding bases and dimensions.
- Null Space and Column Space: These are specific subspaces associated with a matrix that provide valuable information about the matrix's properties.
- Rank-Nullity Theorem: This theorem relates the rank and nullity of a matrix, providing a powerful tool for analyzing linear systems.
- Eigenvalues and Eigenvectors: These concepts are crucial for understanding the behavior of linear transformations and the stability of systems.
By delving into these topics, you will develop a more comprehensive understanding of linear algebra and its applications. Remember, the key to mastering these concepts is practice. Work through various examples and problems to solidify your understanding and build your problem-solving skills.
