Problem 27 - Differential Equations and Linear Algebra

Question

Determine all eigenvalues and corresponding eigenvectors of the given matrix. $$\left[\begin{array}{rrr}-2 & 1 & 0 \\\1 & -1 & -1 \\\1 & 3 & -3\end{array}\right].$$

Step-by-Step Solution

Answer

The eigenvalues of the given matrix are 0, -1, and 3. Their corresponding eigenvectors (up to a nonzero scalar multiple) are \(\begin{bmatrix}2\\1\\1\end{bmatrix}\), \(\begin{bmatrix}1\\1\\1\end{bmatrix}\), and \(\begin{bmatrix}\dfrac{1}{15}\\\dfrac{1}{3}\\1\end{bmatrix}\).

Step 1: Calculate the characteristic polynomial

First, we will find the characteristic polynomial of the matrix. The characteristic polynomial is given by the determinant of (A - λI), where A is the given matrix, λ is the eigenvalue, and I is the identity matrix. We will use the matrix $$\left[\begin{array}{rrr}-2-\lambda & 1 & 0 \\\1 & -1-\lambda & -1 \\\1 & 3 & -3-\lambda\end{array}\right].$$ Now, compute the determinant: \begin{align*} \text{det}(A-\lambda I) = \begin{vmatrix} -2-\lambda & 1 & 0 \\ 1 & -1-\lambda & -1 \\ 1 & 3 & -3-\lambda \end{vmatrix} &= (-2-\lambda) ((-1-\lambda)(-3-\lambda) - (3)(-1)) - (1)(-1(-3-\lambda)) \\ &= (-\lambda^3 + 4\lambda^2 + 3\lambda). \end{align*}

Step 2: Find the eigenvalues

To find the eigenvalues, we need to solve the equation det(A - λI) = 0 for λ. In our case, the equation is: $$-\lambda^3 + 4\lambda^2 + 3\lambda = 0.$$ The polynomial can be factored as follows: $$-\lambda(\lambda^2 - 4\lambda - 3) = 0.$$ This equation has three solutions: λ = 0, λ = -1, and λ = 3. These are our eigenvalues.

Step 3: Find the eigenvectors corresponding to each eigenvalue

Now, for each eigenvalue, we will substitute it into the equation (A - λI)v = 0 and solve for the eigenvectors. 1. For λ = 0, we have: $$\left[\begin{array}{rrr} -2 & 1 & 0 \\ 1 & -1 & -1 \\ 1 & 3 & -3 \end{array}\right] \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}.$$ Row reduce the augmented matrix to find the eigenvector: \( \left[\begin{array}{ccc | c} -2 & 1 & 0 & 0\\ 1 & -1 & -1 & 0 \\ 1 & 3 & -3 & 0 \end{array}\right] \sim\\\) \( \left[\begin{array}{ccc | c} 1 & -1 & -1 & 0\\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]. \) Thus, we have x_1 = x_2 + x_3 and x_2 = x_3. Let x_3 = t; then x_1 = 2t, x_2 = t. The eigenvector for λ = 0 is: \(\begin{bmatrix}2t\\t\\t\end{bmatrix} = t\begin{bmatrix}2\\1\\1\end{bmatrix}.\) 2. For λ = -1, substitute it into (A - λI)v = 0: $$\left[\begin{array}{rrr} -1 & 1 & 0 \\ 1 & 0 & -1 \\ 1 & 3 & -2 \end{array}\right] \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}.$$ Row reduce the augmented matrix: \( \left[\begin{array}{ccc | c} -1 & 1 & 0 & 0\\ 1 & 0 & -1 & 0 \\ 1 & 3 & -2 & 0 \end{array}\right] \sim\\\) \( \left[\begin{array}{ccc | c} 1 & -1 & 0 & 0\\ 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]. \) We have x_1 = x_2 and x_2 = x_3. Let x_3 = t; then x_1 = t, x_2 = t. The eigenvector for λ = -1 is: \(\begin{bmatrix}t\\t\\t\end{bmatrix} = t\begin{bmatrix}1\\1\\1\end{bmatrix}.\) 3. For λ = 3, substitute it into (A - λI)v = 0: $$\left[\begin{array}{rrr} -5 & 1 & 0 \\ 1 & -4 & -1 \\ 1 & 3 & -6 \end{array}\right] \begin{bmatrix}x_1\\x_2\\x_3\end{bmatrix} = \begin{bmatrix}0\\0\\0\end{bmatrix}.$$ Row reduce the augmented matrix: \( \left[\begin{array}{ccc | c} -5 & 1 & 0 & 0\\ 1 & -4 & -1 & 0 \\ 1 & 3 & -6 & 0 \end{array}\right] \sim\\\) \( \left[\begin{array}{ccc | c} 1 & -\frac{1}{5} & 0 & 0\\ 0 & 1 & -\frac{1}{3} & 0 \\ 0 & 0 & 1 & 0 \end{array}\right]. \) Thus, x_1 = \(\dfrac{1}{5}\)x_2, x_2 = \(\dfrac{1}{3}\)x_3, and x_3 is free. Let x_3 = t; then x_1 = \(\dfrac{1}{5}\cdot\dfrac{1}{3}t\) = \(\dfrac{1}{15}t\), x_2 = \(\dfrac{1}{3}t\). The eigenvector for λ = 3 is: \(\begin{bmatrix}\dfrac{1}{15}t\\\dfrac{1}{3}t\\t\end{bmatrix} = t\begin{bmatrix}\dfrac{1}{15}\\\dfrac{1}{3}\\1\end{bmatrix}.\) In conclusion, the eigenvalues are 0, -1, and 3, and their corresponding eigenvectors are (up to a nonzero scalar multiple) \(\begin{bmatrix}2\\1\\1\end{bmatrix}\), \(\begin{bmatrix}1\\1\\1\end{bmatrix}\), and \(\begin{bmatrix}\dfrac{1}{15}\\\dfrac{1}{3}\\1\end{bmatrix}\).

Key Concepts

Characteristic Polynomial

Understanding the concept of the characteristic polynomial is crucial for determining eigenvalues of a matrix. When given a square matrix \( A \), the characteristic polynomial is formulated as the determinant of the matrix \( A - \lambda I \), where \( \lambda \) is a scalar, and \( I \) is the identity matrix of the same dimensions as \( A \).

This polynomial is of particular interest in linear algebra because its roots, called eigenvalues, describe the scaling factor by which the matrix leaves its eigenvectors unaffected, aside from scaling.

In essence, finding the characteristic polynomial involves:

• Forming the matrix \( A - \lambda I \)
• Calculating the determinant of this new matrix

The solutions to the resulting equation provide the eigenvalues, fundamental in matrix operations and transformations.

Row Reduction

Row reduction is a process used in linear algebra to simplify a matrix to row-echelon form or even reduced row-echelon form. This method is vital for solving systems of linear equations, analyzing matrix rank, and importantly, finding eigenvectors when given eigenvalues.

During the process, we perform elementary row operations that include:

• Swapping two rows
• Multiplying a row by a non-zero scalar
• Adding or subtracting a scalar multiple of one row to another row

After determining the eigenvalues from the characteristic polynomial, row reduction helps in uncovering the corresponding eigenvectors by simplifying the system of equations expressed as \((A - \lambda I)v = 0\). This linear system represents a set of vectors that remain invariant (subject to scaling) under the original matrix transformation.

Matrix Determinant

The determinant of a matrix is a scalar value that is a property of square matrices. It reveals significant information regarding the nature of the matrix including properties such as invertibility. Calculating the determinant is a key technique in linear algebra, especially when determining the characteristic polynomial.

A few critical insights into determinants include:

• A matrix with a determinant of zero is singular, meaning it lacks an inverse.
• The determinant of a matrix can be calculated through cofactor expansion or row reduction techniques for larger matrices.

While finding eigenvalues, we set the determinant of \( A - \lambda I \) to zero to solve for \( \lambda \), essentially leading us to the characteristic equation whose solutions, the roots, are the eigenvalues.

Linear Algebra

Linear Algebra is a field of mathematics revolving around vectors, vector spaces, linear transformations, and the understanding of systems of linear equations. Key operations often involve matrices, and concepts like eigenvalues and eigenvectors are foundational elements.

In essence, linear algebra provides the tools and frameworks necessary to handle multidimensional data efficiently and is applicable in various fields including computer science, physics, and engineering. By understanding the matrix structure and operations such as matrix multiplication, determinants, and row reduction, practitioners can solve complex systems and analyze transformations that describe numerous real-world phenomena.

The interconnectedness of concepts like the characteristic polynomial, eigenvalues, and eigenvectors illustrates how linear algebra forms a coherent system. This synergy facilitates advancements in computations and theoretical developments across multiple disciplines.