Problem 25
Question
Deal with the eigenvalue/eigenvector problem for \(n \times n\) real skew- symmetric matrices. Determine all eigenvalues and corresponding eigenvectors of the matrix $$ A=\left[\begin{array}{rrr} 0 & 4 & -4 \\ -4 & 0 & -2 \\ 4 & 2 & 0 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
For the given skew-symmetric matrix \(A=\left[\begin{array}{rrr} 0 & 4 & -4 \\ -4 & 0 & -2 \\ 4 & 2 & 0 \end{array}\right]\), we found that the real eigenvalue is \(\lambda = 0\), and the corresponding real eigenvector is \(v_0 = \left[\begin{array}{c} 1/2 \\ 1 \\ 1 \end{array}\right]\). The other eigenvalues of skew-symmetric matrices are complex conjugate pairs, which do not have a valid interpretation in the context of real skew-symmetric matrices.
1Step 1: Compute the characteristic equation
For matrix A, we first find (A - λI) by subtracting λ from each diagonal element. Then, we calculate the determinant of the resulting matrix.
\( (A - λI) = \left[\begin{array}{ccc}
0-λ & 4 & -4 \\
-4 & 0-λ & -2 \\
4 & 2 & 0-λ
\end{array}\right] \)
Now, calculate the determinant:
\( |A - λI| = (0-λ) \left(((0-λ)(0-λ)) - (-2)(2)\right) - 4 \left((-4)(2) - (0-λ)(-4)\right) - 4 \left((-4)(-2)- (4)(2)\right)\)
2Step 2: Solve for eigenvalues λ
Simplify the expression obtained in Step 1:
\( |A - λI| = -(λ^3 + 8λ) \)
This is the characteristic equation that we need to solve to find the eigenvalues. Set the expression equal to 0 and solve for λ:
\( -(λ^3 + 8λ) = 0 \)
\( λ^3 + 8λ = 0 \)
\( λ(λ^2 + 8) = 0 \)
From this equation, we can see that λ = 0 is an eigenvalue. Furthermore, the other two eigenvalues will be complex conjugate pairs, as the remaining equation is a quadratic. There are three real eigenvalues for skew-symmetric matrices: {0, λ, -λ}.
3Step 3: Find the eigenvectors corresponding to λ = 0
Now, we will find the eigenvectors corresponding to the eigenvalue λ = 0. We need to solve the system of linear equations given by (A - λI)v = 0:
\(
\left[\begin{array}{ccc}
0 & 4 & -4 \\
-4 & 0 & -2 \\
4 & 2 & 0
\end{array}\right]
\left[\begin{array}{c}
x \\
y \\
z
\end{array}\right] =
\left[\begin{array}{c}
0 \\
0 \\
0
\end{array}\right]
\)
Observe that the first equation, 4y - 4z = 0, becomes y = z after simplification. Therefore, we can choose z = 1, which gives y = 1. Then, from the other two equations, we have x = 1/2. Therefore, the eigenvector corresponding to eigenvalue λ = 0 is:
\( v_0 = \left[\begin{array}{c}
1/2 \\
1 \\
1
\end{array}\right]\)
As we found out in step 2, the other eigenvalues would be complex conjugate pairs. However, since we are working with real skew-symmetric matrices, the complex eigenvectors don't have a valid interpretation in this context. So, we can stop here and conclude that for the given skew-symmetric matrix A, the real eigenvalue is λ = 0, and the corresponding eigenvector is \( v_0 = \left[\begin{array}{c} 1/2 \\ 1 \\ 1 \end{array}\right]\).
Key Concepts
Characteristic EquationSkew-Symmetric MatrixDeterminantSystem of Linear Equations
Characteristic Equation
The characteristic equation is a pivotal tool in linear algebra used to determine the eigenvalues of a matrix. It is derived from the determinant of the matrix when it is subtract from an identity matrix scaled by a variable, commonly denoted as \( \lambda \), representing the eigenvalues. For a given square matrix \( A \), the characteristic equation is represented as \( |A - \lambda I| = 0 \), where \( I \) is the identity matrix of the same dimension as \( A \).
When the determinant is calculated, it results in a polynomial in terms of \( \lambda \). The roots of this polynomial are the eigenvalues of matrix \( A \). In the given exercise, the characteristic equation is found by setting up \( |A - \lambda I| = -(\lambda^3 + 8\lambda) = 0 \), which upon simplification gives us the eigenvalues of the matrix. Understanding how to derive and solve the characteristic equation is essential for finding not just the eigenvalues but also the eigenvectors associated with a matrix.
When the determinant is calculated, it results in a polynomial in terms of \( \lambda \). The roots of this polynomial are the eigenvalues of matrix \( A \). In the given exercise, the characteristic equation is found by setting up \( |A - \lambda I| = -(\lambda^3 + 8\lambda) = 0 \), which upon simplification gives us the eigenvalues of the matrix. Understanding how to derive and solve the characteristic equation is essential for finding not just the eigenvalues but also the eigenvectors associated with a matrix.
Skew-Symmetric Matrix
A skew-symmetric matrix, also known as an antisymmetric matrix, is a square matrix that is equal to the negation of its transpose, meaning \( A = -A^T \). Skew-symmetry implies that all the diagonal elements of such a matrix must be zero and the off-diagonal elements are negatives of each other across the main diagonal.
In the context of eigenvalues, an interesting property of real skew-symmetric matrices is that their eigenvalues are either zero or purely imaginary numbers. For example, in our exercise, the given matrix \( A \) is skew-symmetric, and solving its characteristic equation yields eigenvalues that are real or occur in complex conjugate pairs. It's particularly noteworthy that these matrices always have the eigenvalue zero, which we also observe in the exercise's solution. This inherent symmetry tends to simplify many calculations and offers insights into the geometric transformations described by the matrix.
In the context of eigenvalues, an interesting property of real skew-symmetric matrices is that their eigenvalues are either zero or purely imaginary numbers. For example, in our exercise, the given matrix \( A \) is skew-symmetric, and solving its characteristic equation yields eigenvalues that are real or occur in complex conjugate pairs. It's particularly noteworthy that these matrices always have the eigenvalue zero, which we also observe in the exercise's solution. This inherent symmetry tends to simplify many calculations and offers insights into the geometric transformations described by the matrix.
Determinant
The determinant is a scalar value that is computed from the elements of a square matrix. It is a fundamental quantity in linear algebra, providing information on matrix properties such as invertibility: a matrix is invertible if and only if its determinant is non-zero. The determinant also carries geometric significance, reflecting the scaling factor of the linear transformation associated with the matrix, and the orientation it induces on volume.
When calculating the characteristic equation, the determinant of \( A - \lambda I \) is evaluated. It serves as the engine behind finding eigenvalues as it must be set to zero. The determinant of the skew-symmetric matrix in our exercise yields the polynomial equation \( |A - \lambda I| = -(\lambda^3 + 8\lambda) \), which then gets factored and solved to find the eigenvalues. Mastering the calculation of determinants is crucial, as it is prevalent in understanding matrix-related concepts and solving various problems in algebra.
When calculating the characteristic equation, the determinant of \( A - \lambda I \) is evaluated. It serves as the engine behind finding eigenvalues as it must be set to zero. The determinant of the skew-symmetric matrix in our exercise yields the polynomial equation \( |A - \lambda I| = -(\lambda^3 + 8\lambda) \), which then gets factored and solved to find the eigenvalues. Mastering the calculation of determinants is crucial, as it is prevalent in understanding matrix-related concepts and solving various problems in algebra.
System of Linear Equations
A system of linear equations consists of two or more equations with a set of variables. The goal is to find values for the variables that satisfy all equations in the system simultaneously. When it comes to finding eigenvectors, we solve a system where the matrix equation \( (A - \lambda I)v = 0 \) represents the system, and \( v \) is the eigenvector associated with the eigenvalue \( \lambda \).
In our exercise, setting \( \lambda \) to zero transforms the problem into solving a homogeneous system of linear equations to find the eigenvector corresponding to the eigenvalue lambda equals zero. By performing row operations or using various methods like substitution or matrix reduction, we identify the eigenvector \( v_0 = \left[\begin{array}{c} 1/2 \ 1 \ 1 \end{array}\right] \). Understanding how to navigate through systems of linear equations is vital for unlocking the relationship between a matrix and its eigenproperties, which is a cornerstone in linear algebra and its applications.
In our exercise, setting \( \lambda \) to zero transforms the problem into solving a homogeneous system of linear equations to find the eigenvector corresponding to the eigenvalue lambda equals zero. By performing row operations or using various methods like substitution or matrix reduction, we identify the eigenvector \( v_0 = \left[\begin{array}{c} 1/2 \ 1 \ 1 \end{array}\right] \). Understanding how to navigate through systems of linear equations is vital for unlocking the relationship between a matrix and its eigenproperties, which is a cornerstone in linear algebra and its applications.
Other exercises in this chapter
Problem 25
Find the Jordan canonical form \(J\) for the matrix \(A_{1}\) and determine an invertible matrix \(S\) such that \(S^{-1} A S=J\). \(A=\left[\begin{array}{rrr}2
View solution Problem 25
Let \(A\) be a nondefective matrix. Then $$S^{-1} A S=D,$$ where \(D\) is a diagonal matrix. This can be written as $$A=S D S^{-1},$$ Use this result to show th
View solution Problem 25
Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{rrr}0 & 1 & -1 \\\0 & 2 & 0 \\\2 & -1 & 3\end{array}\right]$
View solution Problem 25
Determine a basis for each eigenspace of \(A\) and sketch the eigenspaces. $$A=\left[\begin{array}{ll} 2 & 3 \\ 0 & 2 \end{array}\right]$$
View solution