Problem 27 - Differential Equations and Linear Algebra

Question

Determine a basis for each eigenspace of \(A\) and sketch the eigenspaces. $$\begin{aligned} &A=\left[\begin{array}{rrr} 3 & 1 & -1 \\ 1 & 3 & -1 \\ -1 & -1 & 3 \end{array}\right]\\\ &\text { characteristic polynomial } p(\lambda)=(5-\lambda)(\lambda-2)^{2} \end{aligned}$$

Step-by-Step Solution

Answer

The eigenspaces of the matrix A with the given eigenvalues \(\lambda_1 = 5\) and \(\lambda_2 = 2\) have the following bases: \[ E_{\lambda_1} = \left\{\left[\begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right]\right\},\quad E_{\lambda_2} = \left\{\left[\begin{array}{r} -1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 1 \\ 0 \\ 1 \end{array}\right]\right\} \] The eigenspace for \(\lambda_1\) is a line in 3D space passing through the origin with the direction vector \(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\), while the eigenspace for \(\lambda_2\) is a plane in 3D space that passes through the origin, with basis vectors \(\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\) spanning the eigenspace.

Step 1: Identify the eigenvalues

The given characteristic polynomial is \(p(\lambda) = (5-\lambda)(\lambda-2)^2\), thus we have two eigenvalues: \(\lambda_1 = 5\) and \(\lambda_2 = 2\).

Step 2: Determine the eigenspace for \(\lambda_1\)

For \(\lambda_1 = 5\), we have to find the null space of (A - 5I): \[ (A - 5I) = \left[\begin{array}{rrr} -2 & 1 & -1 \\ 1 & -2 & -1 \\ -1 & -1 & -2 \end{array}\right] \] Row reduction gives us: \[ \left[\begin{array}{rrr} 1 & -1 & 1 \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{array}\right] \] Now, we can write the solutions in parametric form with the free variable z: \[ \begin{cases} x = z \\ y = z \\ z = z \end{cases} \] The eigenspace corresponding to \(\lambda_1 = 5\) has the following basis: \[ E_{\lambda_1} = \left\{\left[\begin{array}{r} 1 \\ 1 \\ 1 \end{array}\right]\right\} \]

Step 3: Determine the eigenspace for \(\lambda_2\)

For \(\lambda_2 = 2\), we have to find the null space of (A - 2I): \[ (A - 2I) = \left[\begin{array}{rrr} 1 & 1 & -1 \\ 1 & 1 & -1 \\ -1 & -1 & 1 \end{array}\right] \] Row reduction gives us: \[ \left[\begin{array}{rrr} 1 & 1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right] \] Now, we can write the solutions in parametric form with the free variables y and z: \[ \begin{cases} x = -y + z \\ y = y \\ z = z \end{cases} \] The eigenspace corresponding to \(\lambda_2 = 2\) has the following basis: \[ E_{\lambda_2} = \left\{\left[\begin{array}{r} -1 \\ 1 \\ 0 \end{array}\right],\left[\begin{array}{r} 1 \\ 0 \\ 1 \end{array}\right]\right\} \]

Step 4: Describe the eigenspaces

The eigenspace for \(\lambda_1 = 5\) is a line in 3D space passing through the origin with the direction vector \(\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}\). The eigenspace for \(\lambda_2 = 2\) is a plane in 3D space that passes through the origin, with the basis vectors \(\begin{bmatrix} -1 \\ 1 \\ 0 \end{bmatrix}\) and \(\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}\) spanning the eigenspace.

Key Concepts

Linear Algebra

Linear algebra is a branch of mathematics that focuses on vectors, vector spaces, and linear transformations between these spaces. It is a fundamental tool used in various fields like engineering, physics, computer science, and economics.

In the context of this exercise, we are dealing with matrices and their properties. A matrix is a rectangular array of numbers arranged in rows and columns. The particular matrix given here is a 3x3 matrix, which is a common type found in linear algebra problems.

The process of finding eigenvalues and their corresponding eigenspaces involves operations like finding the determinant of a matrix and performing row reductions. These are core techniques in linear algebra that allow us to solve for unknowns and understand how transformations work in multi-dimensional spaces.

• **Vectors**: Basic building blocks in linear algebra. Vectors in this case are 3D column matrices representing direction and magnitude.
• **Vector Spaces**: Collections of vectors that can be added together and scaled. Eigenspaces are examples of vector spaces.
• **Matrix Operations**: Includes addition, subtraction, and multiplication. Finding eigenspaces requires you to perform matrix subtraction and row reduction.

Eigenvalues

Eigenvalues are special numbers that provide insights into the behavior of a matrix when it represents a linear transformation. Simply put, for a given transformation matrix \(A\), an eigenvalue \(\lambda\) is a scalar that, when multiplied by an eigenvector, results in a vector that is only scaled and not altered in direction by the transformation.

To find the eigenvalues of a matrix, we typically solve the characteristic polynomial, which in this exercise is given as \((5-\lambda)(\lambda-2)^2\). This polynomial helps us determine the values of \(\lambda\) that satisfy the equation \(\det(A - \lambda I) = 0\), where \(I\) is the identity matrix. Here, the eigenvalues are \(\lambda_1 = 5\) and \(\lambda_2 = 2\).

Understanding eigenvalues allows mathematicians and engineers to discern important properties of transformations, such as stability, oscillations, and resonance in systems.

• **Characteristic Polynomial**: The equation used to find eigenvalues, formed from the determinant of \(A - \lambda I\).
• **Determinant**: A scalar value that is a function of a matrix, significant in finding solutions to systems of linear equations and in determining eigenvalues.
• **Identity Matrix \(I\)**: A diagonal matrix with ones on the main diagonal, utilized in forming \(A - \lambda I\).

Basis

In linear algebra, a basis is a set of vectors in a vector space such that every vector in the space can be expressed as a linear combination of the basis vectors. For eigenspaces, the basis provides the minimal set of vectors necessary to represent all vectors in that space.

For the exercise's first eigenvalue, \(\lambda_1 = 5\), the basis of the eigenspace is the single vector \(\begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix}\), forming a line through the origin in the 3D space. This means that any vector in this eigenspace can be scaled versions of this basis vector.

For the second eigenvalue, \(\lambda_2 = 2\), the basis consists of two vectors: \(\begin{bmatrix} -1 \ 1 \ 0 \end{bmatrix}\) and \(\begin{bmatrix} 1 \ 0 \ 1 \end{bmatrix}\). These vectors form a plane through the origin. Every vector in this plane is a combination of these two basis vectors, allowing us to describe the entire eigenspace with just these vectors.

• **Linear Combinations**: A sum of scalar multiples of vectors, used to express any vector in the space using basis vectors.
• **Span**: The set of all possible linear combinations of a given set of vectors. A basis spans a vector space.
• **Dimension**: The number of vectors in a basis for the space, indicating the minimum number of coordinates needed to specify any vector in that space.