Problem 25
Question
Determine a basis for each eigenspace of \(A\) and sketch the eigenspaces. $$A=\left[\begin{array}{ll} 2 & 3 \\ 0 & 2 \end{array}\right]$$
Step-by-Step Solution
Verified Answer
The eigenvalues of matrix \(A = \left[\begin{array}{cc} 2 & 3 \\ 0 & 2 \end{array}\right]\) is λ = 2. The eigenspace corresponding to this eigenvalue has a basis {v}:
Basis = { \(\left[\begin{array}{c} 1 \\ 0 \end{array}\right] \)}
The eigenspace is a subspace that only consists of the x-axis and can be sketched by drawing a horizontal line at \(y = 0\).
1Step 1: Find the eigenvalues of matrix A
To find the eigenvalues of matrix A, we first need to find the determinant of (A - λI) where λ is an eigenvalue, and I is the identity matrix.
\(A-\lambda I=\left[\begin{array}{cc}
2-\lambda & 3 \\
0 & 2-\lambda
\end{array}\right]\)
Now, we find the determinant:
\(|A-\lambda I|= (2-\lambda)(2-\lambda) - (0)(3)\)
\(|A-\lambda I|= (2-\lambda)^2\)
To find the eigenvalues, we set this determinant equation equal to 0:
\((2-\lambda)^2 = 0\)
By solving for λ, we can find the eigenvalues.
2Step 2: Find a basis for each eigenspace
There is only one eigenvalue, λ = 2. To find the eigenspace corresponding to this eigenvalue, we need to find the null space for the matrix (A - λI):
\((A-2 I) = \left[\begin{array}{cc}
0 & 3 \\
0 & 0
\end{array}\right]\)
Now, let's find the solution to the system of homogeneous linear equations formed by the upper-left matrix:
\(3y = 0\)
The general eigenvector that satisfies this equation is:
\(v = \left[\begin{array}{c}
1 \\
0
\end{array}\right]\)
Thus, the eigenspace corresponding to the eigenvalue λ = 2 has a basis {v}:
Basis = { \(\left[\begin{array}{c}
1 \\
0
\end{array}\right] \)}
3Step 3: Sketch the eigenspaces.
Since there is only one eigenspace corresponding to the eigenvalue λ = 2, with a basis vector:
\(\left[\begin{array}{c}
1 \\
0
\end{array}\right] \)
The eigenspace is a subspace that only consists of the x-axis. This can be easily sketched by drawing an x-axis line (horizontal line) at \(y = 0\).
Key Concepts
EigenvaluesLinear AlgebraNull SpaceHomogeneous Equations
Eigenvalues
Understanding eigenvalues is fundamental in linear algebra and has various applications in fields such as physics, engineering, and computer science. An eigenvalue, denoted by \(\text{\lambda}\), is a special scalar associated with a square matrix \(A\). It represents a value for which there exists a non-zero vector \(v\), called an eigenvector, that when multiplied by the matrix does not change direction but may be scaled by \(\text{\lambda}\).
To find eigenvalues, we solve the characteristic equation \(\text{det}(A-\lambda I) = 0\), where \(I\) is the identity matrix of the same size as A. The solutions to this equation give us the eigenvalues of matrix \(A\). In the given exercise, there is one eigenvalue \(\lambda = 2\). In matrices that possess distinct eigenvalues, the eigenvectors corresponding to each eigenvalue are linearly independent, leading to a basis for the eigenspace.
To find eigenvalues, we solve the characteristic equation \(\text{det}(A-\lambda I) = 0\), where \(I\) is the identity matrix of the same size as A. The solutions to this equation give us the eigenvalues of matrix \(A\). In the given exercise, there is one eigenvalue \(\lambda = 2\). In matrices that possess distinct eigenvalues, the eigenvectors corresponding to each eigenvalue are linearly independent, leading to a basis for the eigenspace.
Linear Algebra
Linear algebra is a branch of mathematics focused on vector spaces and linear mappings between these spaces. It deals with vectors, matrices, and systems of linear equations. The concepts of eigenvalues and eigenvectors arise naturally within the study of linear transformations. In the context of the exercise, we use linear algebra to understand how the matrix \(A\) transforms vectors in its space and to characterize the eigenspaces of \(A\).
Knowing the eigenvalues, we can investigate the geometric properties of these transformations, such as scaling and rotating vectors in space. The relevance of linear algebra transcends pure mathematics, providing key insights into areas such as quantum mechanics, computer graphics, and machine learning algorithms.
Knowing the eigenvalues, we can investigate the geometric properties of these transformations, such as scaling and rotating vectors in space. The relevance of linear algebra transcends pure mathematics, providing key insights into areas such as quantum mechanics, computer graphics, and machine learning algorithms.
Null Space
The null space (or kernel) of a matrix \(A\) refers to a set of vectors that, when multiplied by \(A\), result in the zero vector. This concept is directly related to solving the equation \(Ax = 0\), where the solution set forms the basis of the null space of \(A\).
In our exercise, the null space is required to find the eigenspace for the eigenvalue \(\lambda = 2\). To find the eigenspace, we compute the null space of \(A - \lambda I\). This involves solving the homogeneous system of linear equations given by the matrix equation \((A - \lambda I)x = 0\). The basis vectors obtained from the null space define the directions in which the matrix \(A\) scales vectors by the factor \(\lambda\).
In our exercise, the null space is required to find the eigenspace for the eigenvalue \(\lambda = 2\). To find the eigenspace, we compute the null space of \(A - \lambda I\). This involves solving the homogeneous system of linear equations given by the matrix equation \((A - \lambda I)x = 0\). The basis vectors obtained from the null space define the directions in which the matrix \(A\) scales vectors by the factor \(\lambda\).
Homogeneous Equations
Homogeneous equations are linear equations of the form \(Ax = 0\), where the zero on the right side represents a vector composed of all zeros. These types of equations are pivotal when finding the null space of a matrix and are crucial in our exercise for determining the eigenspace's basis.
The solutions to homogeneous equations are important because they form a vector space—a collection of vectors that can be scaled and added together while still being part of the set. When dealing with homogeneous equations, if we find one non-trivial (non-zero) solution, we infinitely have many, as any scalar multiple of that solution is also a solution. These solutions help define the structure of the matrix’s null space, and by extension, the eigenspaces associated with its eigenvalues.
The solutions to homogeneous equations are important because they form a vector space—a collection of vectors that can be scaled and added together while still being part of the set. When dealing with homogeneous equations, if we find one non-trivial (non-zero) solution, we infinitely have many, as any scalar multiple of that solution is also a solution. These solutions help define the structure of the matrix’s null space, and by extension, the eigenspaces associated with its eigenvalues.
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