Question

Find the eigenvalues of the given matrix. For each eigenvalue, give an eigenvector. $$\left[\begin{array}{ccc}1 & -2 & -3 \\ 0 & 3 & 0 \\ 0 & -1 & -1\end{array}\right]$$

Short answer

Eigenvalues: 1, -3, -1; Eigenvectors: \(\begin{pmatrix}3\\0\\1\end{pmatrix}, \begin{pmatrix}3\\0\\4\end{pmatrix}, \begin{pmatrix}3\\0\\2\end{pmatrix}\).

Step-by-step solution

Understand the Problem

We are given a matrix and need to find its eigenvalues and eigenvectors. Eigenvalues are values of \( \lambda \) that satisfy the equation \( A - \lambda I = 0 \). For each eigenvalue, we must find a corresponding eigenvector.

Write Characteristic Equation

To find the eigenvalues of the given matrix, we calculate the characteristic equation \( |A - \lambda I| = 0 \), where \( A \) is the given matrix and \( I \) is the identity matrix.\[ A = \begin{pmatrix} 1 & -2 & -3 \ 0 & 3 & 0 \ 0 & -1 & -1 \end{pmatrix}, \quad I = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} \]Hence, \[ A - \lambda I = \begin{pmatrix} 1-\lambda & -2 & -3 \ 0 & 3-\lambda & 0 \ 0 & -1 & -1-\lambda \end{pmatrix} \]Calculate the determinant of this matrix and set it to zero.

Solve for Eigenvalues

Find the determinant of \( A - \lambda I \).\[ |A - \lambda I| = \begin{vmatrix} 1-\lambda & -2 & -3 \ 0 & 3-\lambda & 0 \ 0 & -1 & -1-\lambda \end{vmatrix} = (1-\lambda)((3-\lambda)(-1-\lambda) - 0(0)) \]Simplify the determinant:\[ = (1-\lambda)((\lambda + 3)(\lambda + 1)) \]The characteristic equation is \((1-\lambda)((\lambda + 3)(\lambda + 1)) = 0\). Solve this to get eigenvalues: \( \lambda_1 = 1 \), \( \lambda_2 = -3 \), and \( \lambda_3 = -1 \).

Find Eigenvectors for \( \lambda_1 = 1 \)

Substitute \( \lambda = 1 \) into \( A - \lambda I \):\[ A - I = \begin{pmatrix} 0 & -2 & -3 \ 0 & 2 & 0 \ 0 & -1 & -2 \end{pmatrix} \]Convert to an augmented matrix and row-reduce to find the null space, representing the eigenvector. Solving gives an eigenvector of \( \begin{pmatrix} 3 \ 0 \ 1 \end{pmatrix} \).

Find Eigenvectors for \( \lambda_2 = -3 \)

Substitute \( \lambda = -3 \) into \( A - \lambda I \):\[ A + 3I = \begin{pmatrix} 4 & -2 & -3 \ 0 & 6 & 0 \ 0 & -1 & 2 \end{pmatrix} \]Convert to an augmented matrix and row-reduce to find the null space. The eigenvector is \( \begin{pmatrix} 3 \ 0 \ 4 \end{pmatrix} \).

Find Eigenvectors for \( \lambda_3 = -1 \)

Substitute \( \lambda = -1 \) into \( A - \lambda I \):\[ A + I = \begin{pmatrix} 2 & -2 & -3 \ 0 & 4 & 0 \ 0 & -1 & 0 \end{pmatrix} \]Convert to an augmented matrix and row-reduce to find the null space. The eigenvector is \( \begin{pmatrix} 3 \ 0 \ 2 \end{pmatrix} \).

Key concepts

Eigenvectors

Eigenvectors are key concepts in linear algebra, used to understand linear transformations better. They are non-zero vectors that only change by a scalar factor when a linear transformation is applied. If you have a matrix \(A\), and it acts on a vector \(\mathbf{v}\), the vector remains in the same direction, only scaled by the eigenvalue \(\lambda\). This can be expressed by the equation \(A\mathbf{v} = \lambda \mathbf{v}\). This equation signifies that multiplying a matrix by an eigenvector results in the same vector scaled by a certain factor. This property is essential to many applications, from engineering to machine learning, helping to simplify complex systems. To find eigenvectors, you need to first determine the eigenvalues from the characteristic equation, and then solve the system \((A - \lambda I)\mathbf{v} = 0\) by setting it to zero and solving for \(\mathbf{v}\).

Characteristic Equation

The characteristic equation is a polynomial equation in \(\lambda\) that is used to find the eigenvalues of a matrix. It is derived from the determinant of \(A - \lambda I\), where \(A\) is a matrix and \(I\) is the identity matrix of the same order. To obtain the characteristic equation, perform the following steps:

• Subtract \(\lambda\) times the identity matrix from your matrix \(A\).
• Calculate the determinant of the resulting matrix \(A - \lambda I\).
• Equate the determinant to zero, resulting in the characteristic polynomial.

Solving this polynomial equation gives the eigenvalues. For example, if the characteristic equation derived in this problem is \((1-\lambda)((\lambda + 3)(\lambda + 1)) = 0\), solving it reveals the individual eigenvalues: \(\lambda_1 = 1, \lambda_2 = -3, \lambda_3 = -1\). Thus, the characteristic equation is an essential step in finding eigenvalues.

Determinant

The determinant is a special value that can be calculated from a square matrix. It provides important properties regarding the matrix, like whether it is invertible, as a non-zero determinant indicates that the matrix has an inverse. To compute the determinant for a \(3 \times 3\) matrix, like the given matrix with eigenvalues, you can follow:

• Take the first row of the matrix.
• For each element, cross out the row and column to create a \(2 \times 2\) matrix (called the minor of that element).
• Calculate the determinant of each minor, multiply it by the original matrix element and the appropriate sign.

The determinant of a \(3 \times 3\) matrix is often computed using the rule of Sarrus or cofactor expansion as shown in the problem. When finding eigenvalues, you set the determinant of \(A - \lambda I\) to zero, which provides the characteristic equation that helps in solving for eigenvalues.

Matrix Algebra

Matrix algebra involves operations among matrices, which help in representing and solving linear equations. Fundamental operations include addition, subtraction, scalar multiplication, and matrix multiplication. Key points include:

• Matrix addition and subtraction are performed element-wise, requiring matrices of the same dimension.
• Scalar multiplication involves multiplying every element of a matrix by a scalar (a constant).
• Matrix multiplication is more complex and involves the dot product of rows and columns, whereby the number of columns in the first matrix must equal the number of rows in the second.

In terms of eigenvalues and eigenvectors, matrix algebra helps rearrange terms, create the identity matrix, and simplify systems of equations necessary for solving them. For instance, constructing the matrix \(A - \lambda I\) involves matrix subtraction and scalar multiplication—basic yet critical operations in matrix algebra. Understanding these operations forms the basis for more advanced topics in linear algebra.