Problem 15
Question
Determine the multiplicity of each eigenvalue and a basis for each eigenspace of the given matrix \(A\). Hence, determine the dimension of each eigenspace and state whether the matrix is defective or nondefective. $$A=\left[\begin{array}{rrr} 2 & 3 & 0 \\ -1 & 0 & 1 \\ -2 & -1 & 4 \end{array}\right]$$
Step-by-Step Solution
Verified Answer
Eigenvalues and their multiplicities: λ1 = 1 (multiplicity 1), λ2 = 2 (multiplicity 1), λ3 = 3 (multiplicity 1).
Eigenspace bases and dimensions: For λ1, basis {(-3, 0, 2)}, dimension 1; for λ2, basis {(1, 1, 3)}, dimension 1; for λ3, basis {(3, 1, 3)}, dimension 1.
Since the dimensions of the eigenspaces equal the multiplicities of the corresponding eigenvalues, the matrix A is nondefective.
1Step 1: Find the eigenvalues of matrix A
To find the eigenvalues, we need to solve the characteristic equation, given by:
\(det(A - \lambda I) = 0\)
where λ is an eigenvalue and I is the \(3 \times 3\) identity matrix.
Substituting A, we get the characteristic equation:
\(det(\left[\begin{array}{ccc} 2-\lambda & 3 & 0 \\ -1 & -\lambda & 1 \\ -2 & -1 & 4-\lambda \end{array}\right]) = 0\)
Expanding the determinant along the first row, we get:
\((2-\lambda)((-\lambda)(4-\lambda) - (1)(-1)) - (3)(-1(-1) - (-1)(4-\lambda)) = 0\)
=> \((2-\lambda) (\lambda^2 - 4\lambda +3) -3(-\lambda+4) = 0\)
Now, we need to solve this polynomial equation to find the eigenvalues λ.
2Step 2: Solve the polynomial equation for the eigenvalues
Solving the polynomial equation from step 1:
\( (\lambda-1)(\lambda-3)(\lambda-2) = 0\)
This gives us 3 eigenvalues: λ1 = 1, λ2 = 2, and λ3 = 3, and each has a multiplicity of 1.
3Step 3: Find the eigenspaces for each eigenvalue and their dimensions
For each eigenvalue λ, we have to find its eigenspace, the null space of (A - λI), and its dimension.
- For λ1 = 1, find the eigenspace:
\(A - λ_1I = \left[\begin{array}{ccc} 1 & 3 & 0 \\ -1 & -1 & 1 \\ -2 & -1 & 3 \end{array}\right]\)
Row reduce to echelon form:
\(\left[\begin{array}{ccc} 1 & 3 & 0 \\ 0 & 2 & 1 \\ 0 & 0 & 0 \end{array}\right]\)
The null space (eigenspace) is spanned by the vector: \(v_1 = \left[\begin{array}{c} -3 \\ 0 \\ 2 \end{array}\right]\)
The eigenspace for λ1 has a dimension of 1.
- For λ2 = 2, find the eigenspace:
\(A - λ_2I = \left[\begin{array}{ccc} 0 & 3 & 0 \\ -1 & -2 & 1 \\ -2 & -1 & 2 \end{array}\right]\)
Row reduce to echelon form:
\(\left[\begin{array}{ccc} 1 & 2 & -1 \\ 0 & 3 & -1 \\ 0 & 0 & 0 \end{array}\right]\)
The null space (eigenspace) is spanned by the vector: \(v_2 = \left[\begin{array}{c} 1 \\ 1 \\ 3 \end{array}\right]\)
The eigenspace for λ2 has a dimension of 1.
- For λ3 = 3, find the eigenspace:
\(A - λ_3I = \left[\begin{array}{ccc} -1 & 3 & 0 \\ -1 & -3 & 1 \\ -2 & -1 & 1 \end{array}\right]\)
Row reduce to echelon form:
\(\left[\begin{array}{ccc} 1 & -3 & 0 \\ 0 & -3 & 1 \\ 0 & 0 & 0 \end{array}\right]\)
The null space (eigenspace) is spanned by the vector: \(v_3 = \left[\begin{array}{c} 3 \\ 1 \\ 3 \end{array}\right]\)
The eigenspace for λ3 has a dimension of 1.
4Step 4: Determine if the matrix is defective or nondefective
Since each eigenspace of matrix A has a dimension equal to the multiplicity of its corresponding eigenvalue (all dimensions are 1), the matrix A is nondefective.
Key Concepts
Characteristic EquationMultiplicity of EigenvaluesDefective and Nondefective MatricesDimension of Eigenspace
Characteristic Equation
The characteristic equation is a fundamental concept in linear algebra central to understanding eigenvalues. At its core, the characteristic equation is a polynomial equation that we derive from a square matrix. It is given by the equation \(det(A - \lambda I) = 0\), where \(A\) is the matrix in question, \(\lambda\) represents an eigenvalue and \(I\) is the identity matrix of the same order as \(A\).
To find the characteristic equation for a given matrix, we subtract \(\lambda\) times the identity matrix from our original matrix and then calculate the determinant of the resulting matrix. The solutions to the polynomial equation that emerges define the possible eigenvalues of the matrix. Understanding how to solve the characteristic equation effectively is crucial to revealing the nature of a matrix and its eigenvalues.
To find the characteristic equation for a given matrix, we subtract \(\lambda\) times the identity matrix from our original matrix and then calculate the determinant of the resulting matrix. The solutions to the polynomial equation that emerges define the possible eigenvalues of the matrix. Understanding how to solve the characteristic equation effectively is crucial to revealing the nature of a matrix and its eigenvalues.
Multiplicity of Eigenvalues
Multiplicity in the context of eigenvalues can be described as twofold; algebraic and geometric. Algebraic multiplicity refers to the number of times an eigenvalue appears as a root of the characteristic equation. Geometric multiplicity, on the other hand, pertains to the dimension of the eigenspace corresponding to an eigenvalue. It is the number of linearly independent eigenvectors associated with that eigenvalue.
In the given exercise, each eigenvalue \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\) has an algebraic multiplicity of 1, indicating they are distinct. The multiplicity of eigenvalues informs us about the potential complexity of the solution space and contributes to the broader understanding of the matrix's structure. It’s also worth noting that the geometric multiplicity can never exceed the algebraic multiplicity.
In the given exercise, each eigenvalue \(\lambda_1\), \(\lambda_2\), and \(\lambda_3\) has an algebraic multiplicity of 1, indicating they are distinct. The multiplicity of eigenvalues informs us about the potential complexity of the solution space and contributes to the broader understanding of the matrix's structure. It’s also worth noting that the geometric multiplicity can never exceed the algebraic multiplicity.
Defective and Nondefective Matrices
The terms defective and nondefective apply to matrices based on the relationship between an eigenvalue's algebraic and geometric multiplicities. A defective matrix is one for which at least one eigenvalue has geometric multiplicity less than its algebraic multiplicity. This implies that there are fewer linearly independent eigenvectors than expected, which can complicate matters such as diagonalization.
A nondefective matrix, contrastingly, enjoys full alignment between each eigenvalue's algebraic and geometric multiplicities. They have a complete basis of eigenvectors and can be diagonalized without issue. In the provided solution, since each eigenvalue's geometric and algebraic multiplicities match, our matrix \(A\) is nondefective, simplifying the study of its properties.
A nondefective matrix, contrastingly, enjoys full alignment between each eigenvalue's algebraic and geometric multiplicities. They have a complete basis of eigenvectors and can be diagonalized without issue. In the provided solution, since each eigenvalue's geometric and algebraic multiplicities match, our matrix \(A\) is nondefective, simplifying the study of its properties.
Dimension of Eigenspace
An eigenspace is fundamentally a linear subspace formed by all eigenvectors corresponding to a specific eigenvalue, combined with the zero vector. The dimension of the eigenspace equates to the number of linearly independent vectors it contains, which is the same as its geometric multiplicity.
When finding the dimension of the eigenspace, we typically convert the equation \(A - \lambda I\) into its row echelon form or reduced row echelon form to identify the null space. If the dimension of the eigenspace equals the algebraic multiplicity of the eigenvalue, we can confidently assert that our matrix is not lacking in any eigenvectors needed to form a basis for that eigenspace. In the exercise, the dimension of each eigenspace is 1, suggesting they are all one-dimensional linear subspaces.
When finding the dimension of the eigenspace, we typically convert the equation \(A - \lambda I\) into its row echelon form or reduced row echelon form to identify the null space. If the dimension of the eigenspace equals the algebraic multiplicity of the eigenvalue, we can confidently assert that our matrix is not lacking in any eigenvectors needed to form a basis for that eigenspace. In the exercise, the dimension of each eigenspace is 1, suggesting they are all one-dimensional linear subspaces.
Other exercises in this chapter
Problem 15
Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{ll}2 & 0 \\\0 & 2\end{array}\right]$$.
View solution Problem 15
Determine whether the given matrix \(A\) is diagonalizable. Where possible, find a matrix \(S\) such that $$S^{-1} A S=\operatorname{diag}\left(\lambda_{1}, \la
View solution Problem 16
Write down all of the possible Jordan canonical form structures, up to a rearrangement of the blocks, for matrices of the specified type. For each Jordan canoni
View solution Problem 16
Give an example of a \(2 \times 2\) matrix \(A\) that has a generalized eigenvector that is not an eigenvector, and exhibit such a generalized eigenvector.
View solution