Question

Find the eigenvalues of the given matrix. For each eigenvalue, give an eigenvector. $$\left[\begin{array}{ccc}3 & 5 & -5 \\ -2 & 3 & 2 \\ -2 & 5 & 0\end{array}\right]$$

Short answer

Eigenvalues: 5, 1, 0. Eigenvectors: \([1, 1, 1]^T\), \([1, 0, 2]^T\), \([0, 1, 1]^T\).

Step-by-step solution

Eigenvalue Definition

To find the eigenvalues of the matrix \( A \), we solve the characteristic equation \( \text{det}(A - \lambda I) = 0 \), where \( \lambda \) is an eigenvalue and \( I \) is the identity matrix of the same size as \( A \).

Formulate the Characteristic Matrix

Subtract \( \lambda \) times the identity matrix from \( A \): \( A - \lambda I = \begin{bmatrix} 3-\lambda & 5 & -5 \ -2 & 3-\lambda & 2 \ -2 & 5 & -\lambda \end{bmatrix} \).

Calculate the Determinant

Compute the determinant of the matrix \( A - \lambda I \): \[ \text{det}(A - \lambda I) = (3-\lambda)((3-\lambda)(-\lambda) - 10) + 5(-2(-\lambda) - 4) - 5(-10 - (3-\lambda)) = -(\lambda^3 - 6\lambda^2 + 15\lambda - 10) = 0 \].

Solve the Characteristic Equation

The characteristic polynomial \( \lambda^3 - 6\lambda^2 + 15\lambda - 10 = 0 \) can be solved using methods such as factoring, synthetic division, or numerical methods to find \( \lambda = 5 \), \( \lambda = 1 \), and \( \lambda = 0 \).

Find Eigenvectors for Each Eigenvalue

Substitute each eigenvalue back into the equation \( (A - \lambda I)\mathbf{v} = 0 \) to find the corresponding eigenvector \( \mathbf{v} \).

Calculate Eigenvector for \( \lambda = 5 \)

For \( \lambda = 5 \), the system \( \begin{bmatrix} -2 & 5 & -5 \ -2 & -2 & 2 \ -2 & 5 & -5 \end{bmatrix}\mathbf{v} = 0 \) reduces to finding a nontrivial solution. One eigenvector is \( \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \).

Calculate Eigenvector for \( \lambda = 1 \)

For \( \lambda = 1 \), \( (A - I) = \begin{bmatrix} 2 & 5 & -5 \ -2 & 2 & 2 \ -2 & 5 & -1 \end{bmatrix} \). Solving gives one eigenvector \( \begin{bmatrix} 1 \ 0 \ 2 \end{bmatrix} \).

Calculate Eigenvector for \( \lambda = 0 \)

For \( \lambda = 0 \), \( A = \begin{bmatrix} 3 & 5 & -5 \ -2 & 3 & 2 \ -2 & 5 & 0 \end{bmatrix} \mathbf{v} = 0 \) leads to the eigenvector \( \begin{bmatrix} 0 \ 1 \ 1 \end{bmatrix} \).

Key concepts

Characteristic Equation

The characteristic equation is central to finding the eigenvalues of a matrix. It is derived from the equation \( \text{det}(A - \lambda I) = 0 \), where \( A \) is your matrix, \( \lambda \) represents the eigenvalues, and \( I \) is the identity matrix of the same size as \( A \). This equation tells us that for matrix \( A \), the eigenvalues are those values of \( \lambda \) which make the determinant of \( A - \lambda I \) equal to zero. By solving this equation, we get the roots that correspond to the eigenvalues of the matrix.

Essentially, it all starts with forming the characteristic matrix, \( A - \lambda I \), where you subtract \( \lambda \times I \) from your original matrix. This sets the stage for calculating the determinant and finally deriving the characteristic polynomial. The main goal is to solve \( \text{det}(A - \lambda I) = 0 \) to find each eigenvalue.

Eigenvectors

Once you have the eigenvalues, the next step is to find the eigenvectors for each eigenvalue. Eigenvectors are special vectors associated with a particular eigenvalue. They satisfy the equation \( (A - \lambda I)\mathbf{v} = 0 \). Here, \( \mathbf{v} \) is the eigenvector, and it must not be the zero vector.

The process of finding eigenvectors involves substitution of each eigenvalue, \( \lambda \), back into the matrix equation. You then solve for \( \mathbf{v} \). This usually forms a system of linear equations, which can often be solved by row reduction techniques or matrix algebra. Each eigenvalue may have one or many eigenvectors, leading to an entire "eigenspace" associated with each eigenvalue.

Determinant

The concept of the determinant is at the heart of calculating eigenvalues, since it is a value that provides essential information about a matrix. The determinant of a matrix gives insight into various properties such as invertibility and the volume changed by the transformation by the matrix.

In the context of finding eigenvalues, the determinant allows us to form the characteristic equation: \( \text{det}(A - \lambda I) = 0 \). A non-zero determinant (of \( A - \lambda I \)) for any particular \( \lambda \) suggests that there are no solutions \( \mathbf{v} \) other than the trivial one. When the determinant equals zero, it signifies that non-trivial solutions, and thus eigenvectors exist, corresponding to the eigenvalue \( \lambda \). Calculating the determinant carefully enables the solving of equations for eigenvalues.

Characteristic Polynomial

The characteristic polynomial is formed as part of solving the characteristic equation \( \text{det}(A - \lambda I) = 0 \). It is a polynomial expression in \( \lambda \) where the degree equals the size \( n \) of the \( n \times n \) matrix. The roots of this polynomial are the sought-after eigenvalues.

For example, the matrix's characteristic polynomial might appear as \( \lambda^3 - 6\lambda^2 + 15\lambda - 10 = 0 \) when simplified. Solving this polynomial gives you the eigenvalues \( \lambda = 0, 1, 5 \). Often the polynomial is solved by factoring (if possible), or using numerical methods like synthetic division or the quadratic formula, depending on the degree and complexity of the polynomial.

The characteristic polynomial is crucial because it provides a method for revealing the structural properties of a matrix by finding its eigenvalues.