Finding Bases For Eigenspaces Matrix Example
Introduction
In linear algebra, eigenvalues and eigenvectors are fundamental concepts that provide valuable insights into the behavior of linear transformations. Eigenvectors are special vectors that, when multiplied by a matrix, result in a scaled version of themselves, with the scaling factor being the eigenvalue. The set of all eigenvectors corresponding to a particular eigenvalue, along with the zero vector, forms a subspace called the eigenspace. Finding a basis for each eigenspace is a crucial step in understanding the matrix's properties and its associated linear transformation. This article will delve into the process of finding bases for the eigenspaces of a given matrix, using the example matrix:
[ 1 0 0 ]
[ -8 4 -6 ]
[ 8 1 9 ]
We will systematically determine the eigenvalues, construct the corresponding eigenspaces, and identify a basis for each eigenspace. This process will involve concepts such as characteristic equations, matrix null spaces, and linear independence. Understanding how to find bases for eigenspaces is essential for various applications, including solving systems of differential equations, analyzing the stability of systems, and performing matrix diagonalization.
1. Determining the Eigenvalues
The journey to finding bases for eigenspaces begins with determining the eigenvalues of the given matrix. Eigenvalues, denoted by 位 (lambda), are the scalars that satisfy the equation Av = 位v, where A is the matrix, v is the eigenvector, and 位 is the eigenvalue. To find these eigenvalues, we need to solve the characteristic equation, which is derived from the determinant of (A - 位I), where I is the identity matrix. For our matrix:
A = [ 1 0 0 ]
[ -8 4 -6 ]
[ 8 1 9 ]
We first construct the matrix (A - 位I):
A - 位I = [ 1-位 0 0 ]
[ -8 4-位 -6 ]
[ 8 1 9-位 ]
The next step is to compute the determinant of this matrix and set it equal to zero. This determinant will give us a polynomial equation in terms of 位, known as the characteristic equation. Solving this equation will yield the eigenvalues of the matrix A. The determinant of (A - 位I) is calculated as follows:
det(A - 位I) = (1-位) * det([4-位 -6], [1 9-位]) - 0 * det([-8 -6], [8 9-位]) + 0 * det([-8 4-位], [8 1])
= (1-位) * ((4-位)(9-位) - (-6)(1))
= (1-位) * (36 - 4位 - 9位 + 位^2 + 6)
= (1-位) * (位^2 - 13位 + 42)
Setting the determinant equal to zero gives us the characteristic equation:
(1-位)(位^2 - 13位 + 42) = 0
Now, we need to solve this equation for 位. The quadratic term can be factored as:
位^2 - 13位 + 42 = (位 - 6)(位 - 7)
Thus, the characteristic equation becomes:
(1-位)(位 - 6)(位 - 7) = 0
The eigenvalues are the roots of this equation, which are:
位1 = 1, 位2 = 6, 位3 = 7
These eigenvalues are the scaling factors associated with the eigenvectors of the matrix A. Each eigenvalue corresponds to an eigenspace, which we will explore in the next section.
2. Finding the Eigenspaces
Having determined the eigenvalues, the next step is to find the eigenspaces corresponding to each eigenvalue. The eigenspace for a given eigenvalue 位 is the set of all eigenvectors v that satisfy the equation (A - 位I)v = 0, along with the zero vector. In other words, the eigenspace is the null space of the matrix (A - 位I). To find the eigenspace for each eigenvalue, we will substitute each eigenvalue back into the equation (A - 位I)v = 0 and solve for the eigenvector v. This process involves setting up a system of linear equations and finding the solutions, which will form the basis for the eigenspace.
2.1 Eigenspace for 位1 = 1
For 位1 = 1, we substitute this value into the matrix (A - 位I):
A - 1I = [ 1-1 0 0 ] = [ 0 0 0 ]
[ -8 4-1 -6 ] [ -8 3 -6 ]
[ 8 1 9-1 ] [ 8 1 8 ]
Now we need to solve the homogeneous system (A - 1I)v = 0, where v is the eigenvector [x, y, z]^T. This gives us the following system of equations:
0x + 0y + 0z = 0
-8x + 3y - 6z = 0
8x + y + 8z = 0
The first equation is trivial. We can add the second and third equations to eliminate x:
(-8x + 3y - 6z) + (8x + y + 8z) = 0
4y + 2z = 0
2y + z = 0
z = -2y
Now, substitute z = -2y into the third equation:
8x + y + 8(-2y) = 0
8x + y - 16y = 0
8x - 15y = 0
8x = 15y
x = (15/8)y
So, the eigenvector v can be expressed as:
v = [ (15/8)y ]
[ y ]
[ -2y ]
We can factor out y to get a basis vector. Let y = 8 to eliminate the fraction:
v = y * [ 15/8 ] = [ 15 ]
[ 1 ] [ 8 ]
[ -2 ] [ -16 ]
Thus, a basis for the eigenspace corresponding to 位1 = 1 is {[15, 8, -16]^T}.
2.2 Eigenspace for 位2 = 6
For 位2 = 6, we substitute this value into the matrix (A - 位I):
A - 6I = [ 1-6 0 0 ] = [ -5 0 0 ]
[ -8 4-6 -6 ] [ -8 -2 -6 ]
[ 8 1 9-6 ] [ 8 1 3 ]
Now we solve the homogeneous system (A - 6I)v = 0:
-5x + 0y + 0z = 0
-8x - 2y - 6z = 0
8x + y + 3z = 0
From the first equation, we have x = 0. Substituting x = 0 into the second and third equations gives:
-2y - 6z = 0
y + 3z = 0
y + 3z = 0
y = -3z
So, the eigenvector v can be expressed as:
v = [ 0 ]
[ -3z ]
[ z ]
We can factor out z to get a basis vector. Let z = 1:
v = z * [ 0 ] = [ 0 ]
[ -3 ] [ -3 ]
[ 1 ] [ 1 ]
Thus, a basis for the eigenspace corresponding to 位2 = 6 is {[0, -3, 1]^T}.
2.3 Eigenspace for 位3 = 7
For 位3 = 7, we substitute this value into the matrix (A - 位I):
A - 7I = [ 1-7 0 0 ] = [ -6 0 0 ]
[ -8 4-7 -6 ] [ -8 -3 -6 ]
[ 8 1 9-7 ] [ 8 1 2 ]
Now we solve the homogeneous system (A - 7I)v = 0:
-6x + 0y + 0z = 0
-8x - 3y - 6z = 0
8x + y + 2z = 0
From the first equation, we have x = 0. Substituting x = 0 into the second and third equations gives:
-3y - 6z = 0
y + 2z = 0
y + 2z = 0
y = -2z
So, the eigenvector v can be expressed as:
v = [ 0 ]
[ -2z ]
[ z ]
We can factor out z to get a basis vector. Let z = 1:
v = z * [ 0 ] = [ 0 ]
[ -2 ] [ -2 ]
[ 1 ] [ 1 ]
Thus, a basis for the eigenspace corresponding to 位3 = 7 is {[0, -2, 1]^T}.
3. Bases for the Eigenspaces
Having computed the eigenspaces for each eigenvalue, we can now summarize the bases for each eigenspace. A basis for a subspace is a set of linearly independent vectors that span the subspace. In the context of eigenspaces, the basis vectors are eigenvectors that generate the entire eigenspace through linear combinations. We have found the following bases for the eigenspaces corresponding to the eigenvalues 位1 = 1, 位2 = 6, and 位3 = 7:
- Eigenspace for 位1 = 1: The basis is {[15, 8, -16]^T}.
- Eigenspace for 位2 = 6: The basis is {[0, -3, 1]^T}.
- Eigenspace for 位3 = 7: The basis is {[0, -2, 1]^T}.
These bases provide a fundamental understanding of the behavior of the matrix A with respect to these eigenvectors. When A acts on any vector in the eigenspace corresponding to 位1 = 1, the vector is scaled by a factor of 1. Similarly, vectors in the eigenspace for 位2 = 6 are scaled by 6, and vectors in the eigenspace for 位3 = 7 are scaled by 7. These scaling behaviors are crucial in many applications, including the diagonalization of matrices.
4. Conclusion
In conclusion, we have successfully found the bases for the eigenspaces of the given matrix:
[ 1 0 0 ]
[ -8 4 -6 ]
[ 8 1 9 ]
by systematically determining the eigenvalues and then solving for the corresponding eigenvectors. The eigenvalues were found to be 位1 = 1, 位2 = 6, and 位3 = 7. For each eigenvalue, we constructed the matrix (A - 位I) and solved the homogeneous system (A - 位I)v = 0 to find the eigenvectors. This led us to the following bases for the eigenspaces:
- Eigenspace for 位1 = 1: Basis = {[15, 8, -16]^T}
- Eigenspace for 位2 = 6: Basis = {[0, -3, 1]^T}
- Eigenspace for 位3 = 7: Basis = {[0, -2, 1]^T}
This process demonstrates a fundamental aspect of linear algebra, providing insight into how matrices transform vectors in specific directions (eigenvectors) and by what factors (eigenvalues). The ability to find these bases is crucial for various applications in mathematics, physics, engineering, and computer science. Understanding eigenspaces and eigenvalues allows for the simplification of complex problems by transforming them into simpler, diagonal forms, making analysis and computation more manageable. The concepts and techniques discussed in this article are essential tools for anyone working with linear transformations and matrix analysis.
