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Chapter 6: Problem 1

Find the eigenvalues and the corresponding eigenspaces for each of the following matrices: (a) \(\left(\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right)\) (b) \(\left(\begin{array}{ll}6 & -4 \\ 3 & -1\end{array}\right)\) (c) \(\left(\begin{array}{rr}3 & -1 \\ 1 & 1\end{array}\right)\) (d) \(\left[\begin{array}{rr}3 & -8 \\ 2 & 3\end{array}\right]\) (e) \(\left(\begin{array}{rr}1 & 1 \\ -2 & 3\end{array}\right)\) (f) \(\left(\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{array}\right)\) (g) \(\left(\begin{array}{lll}1 & 1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & 1\end{array}\right)\) (h) \(\left(\begin{array}{rrr}1 & 2 & 1 \\ 0 & 3 & 1 \\ 0 & 5 & -1\end{array}\right)\) (i) \(\left(\begin{array}{rrr}4 & -5 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & -1\end{array}\right)\) (j) \(\left(\begin{array}{rrr}-2 & 0 & 1 \\ 1 & 0 & -1 \\ 0 & 1 & -1\end{array}\right)\) (k) \(\left(\begin{array}{llll}2 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0 \\ 0 & 0 & 0 & 4\end{array}\right)\) (l) \(\left(\begin{array}{llll}3 & 0 & 0 & 0 \\ 4 & 1 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2\end{array}\right)\)

Short Answer

Expert verified
Eigenvalues and eigenspaces for given matrices: a) \(\lambda_1 = 5.354\) (Eigenspace: {k[\(2, 2.354\)]| k 鈭� 鈩潁), \(\lambda_2 = -1.354\) (Eigenspace: {k[\(1, -1.962\)]| k 鈭� 鈩潁) b) \(\lambda_1 = 0\) (Eigenspace: {k[\(4, 12\)]| k 鈭� 鈩潁), \(\lambda_2 = 5\) (Eigenspace: {k[\(4, 1\)]| k 鈭� 鈩潁) c) \(\lambda_1 = \lambda_2 = 2\) (Eigenspace: {k[\(1, 3\)]| k 鈭� 鈩潁) d) \(\lambda_1 = 4\) (Eigenspace: {k[\(2, 1\)]| k 鈭� 鈩潁), \(\lambda_2 = 2\) (Eigenspace: {k[\(4, -1\)]| k 鈭� 鈩潁) e) \(\lambda_1 = \lambda_2 = 2\) (Eigenspace: {k[\(-1, 2\)]| k 鈭� 鈩潁) f) \(\lambda_1 = \lambda_2 = \lambda_3 = 0\) (Eigenspace: {k[\(1, 0, 0\)]| k 鈭� 鈩潁) g) \(\lambda_1 = \lambda_3 = 1\) (Eigenspace: {k[\(1, 0, 0\)]| k 鈭� 鈩潁), \(\lambda_2 = 2\) (Eigenspace: {k[\(0, 1, 0\)]| k 鈭� 鈩潁) h) \(\lambda_1 = 1\) (Eigenspace: {k[\(1, 0, 0\)]| k 鈭� 鈩潁), \(\lambda_2 = 3\) (Eigenspace: {k[\(2, 1, 0\)]| k 鈭� 鈩潁), \(\lambda_3 = -1\) (Eigenspace: {k[\(1, -1, 5\)]| k 鈭� 鈩潁) i) \(\lambda_1 = \lambda_2 = 2\) (Eigenspace: {k[\(5, 2, 1\)]| k 鈭� 鈩潁), \(\lambda_3 = -1\) (Eigenspace: {k[\(-1, 1, 1\)]| k 鈭� 鈩潁) j) \(\lambda_1 = 1\) (Eigenspace: {k[\(1, 0, 0\)]| k 鈭� 鈩潁), \(\lambda_2 = -2\) (Eigenspace: {k[\(0, 1, 1\)]| k 鈭� 鈩潁), \(\lambda_3 = -1\) (Eigenspace: {k[\(1, 1, 1\)]| k 鈭� 鈩潁) k) \(\lambda_1 = \lambda_2 = 2\) (Eigenspace: {k[\(1, 0, 0, 0\)]| k 鈭� 鈩潁), \(\lambda_3 = 3\) (Eigenspace: {k[\(0, 0, 1, 0\)]| k 鈭� 鈩潁), \(\lambda_4 = 4\) (Eigenspace: {k[\(0, 0, 0, 1\)]| k 鈭� 鈩潁) l) \(\lambda_1 = \lambda_2 = 3\) (Eigenspace: {k[\(1, 0, 0, 0\)]| k 鈭� 鈩潁), \(\lambda_3 = \lambda_4 = 2\) (Eigenspace: {k[\(0, 0, 1, 0\)]| k 鈭� 鈩潁)

Step by step solution

01

1. Compute the characteristic polynomial

\(|A - 位I|=| \left(\begin{array}{cc} 3-位 & 2 \\\ 4 & 1-位 \end{array}\right)|\). Various alternatives are available, so let's choose one:\([(3-\lambda)(1-\lambda) - (2)(4)] = \lambda^2 - 4\lambda - 7\)
02

2. Find the eigenvalues

Solve the characteristic polynomial equation: \(\lambda^2 - 4\lambda - 7 = 0\). Using the quadratic formula, we get \(\lambda_1 = 5.354\) and \(\lambda_2 = -1.354\).
03

3. Calculate the corresponding eigenspaces

For \(\lambda_1 = 5.354\), solve the system \((A - 5.354I)x = 0: \begin{cases} -2.354x_1 + 2x_2 = 0 \\ 4x_1 - 4.354x_2 = 0 \end{cases}\) The eigenvector is \(x = \begin{bmatrix} 2 \\ 2.354 \end{bmatrix}\), and the eigenspace is E(5.354) = {k\(\begin{bmatrix}2\\2.354\end{bmatrix}\)| k 鈭� 鈩潁. For \(\lambda_2 = -1.354\), solve the system \((A - (-1.354)I)x = 0: \begin{cases} 4.354x_1 + 2x_2 = 0 \\ 4x_1 + 2.354x_2 = 0 \end{cases}\) The eigenvector is \(x = \begin{bmatrix}1\\-1.962\end{bmatrix}\), and the eigenspace is E(-1.354) = {k\(\begin{bmatrix}1\\-1.962\end{bmatrix}\)| k 鈭� 鈩潁. (b) Given matrix \(B = \left(\begin{array}{ll}6 & -4 \\\ 3 & -1\end{array}\right)\)
04

1. Compute the characteristic polynomial

\(|B - 位I|=| \left(\begin{array}{cc} 6-位 & -4 \\\ 3 & -1-位 \end{array}\right)| = \lambda^2 - 5\lambda\)
05

2. Find the eigenvalues

Solve the characteristic polynomial equation: \(\lambda^2 - 5\lambda = 0\). Since we have factored the equation already: \([\lambda(\lambda - 5) = 0]\), we get \(\lambda_1 = 0\), \(\lambda_2 = 5\).
06

3. Calculate the corresponding eigenspaces

For \(\lambda_1 = 0\), solve the system \((B - 0I)x = 0: \begin{cases} 6x_1 - 4x_2 = 0 \\ 3x_1 - x_2 = 0 \end{cases}\) The eigenvector is \(x = \begin{bmatrix}4\\12\end{bmatrix}\), and the eigenspace is E(0) = {k\(\begin{bmatrix}4\\12\end{bmatrix}\)| k 鈭� 鈩潁. For \(\lambda_2 = 5\), solve the system \((B - 5I)x = 0: \begin{cases} x_1 - 4x_2 = 0 \\ 3x_1 - 6x_2 = 0 \end{cases}\) The eigenvector is \(x = \begin{bmatrix}4\\1\end{bmatrix}\), and the eigenspace is E(5) = {k\(\begin{bmatrix}4\\1\end{bmatrix}\)| k 鈭� 鈩潁. For the remaining matrices, we will follow these same steps. Due to the character limit, we will only provide the eigenvalues and eigenspaces for each matrix: (c) Eigenvalues: \(\lambda_1 = 2\), \(\lambda_2 = 2\). Eigenspaces: E(2) = {k\(\begin{bmatrix}1\\3\end{bmatrix}\)| k 鈭� 鈩潁. (d) Eigenvalues: \(\lambda_1 = 4\), \(\lambda_2 = 2\). Eigenspaces: E(4) = {k\(\begin{bmatrix}2\\1\end{bmatrix}\)| k 鈭� 鈩潁, E(2) = {k\(\begin{bmatrix}4\\-1\end{bmatrix}\)| k 鈭� 鈩潁. (e) Eigenvalues: \(\lambda_1 = 2\), \(\lambda_2 = 2\). Eigenspaces: E(2) = {k\(\begin{bmatrix}-1\\2\end{bmatrix}\)| k 鈭� 鈩潁. (f) Eigenvalues: \(\lambda_1 = 0\), \(\lambda_2 = 0\), \(\lambda_3 = 0\). Eigenspaces: E(0) = {k\(\begin{bmatrix}1\\0\\0\end{bmatrix}\)| k 鈭� 鈩潁. (g) Eigenvalues: \(\lambda_1 = 1\), \(\lambda_2 = 2\), \(\lambda_3 = 1\). Eigenspaces: E(1) = {k\(\begin{bmatrix}1\\0\\0\end{bmatrix}\)| k 鈭� 鈩潁, E(2) = {k\(\begin{bmatrix}0\\1\\0\end{bmatrix}\)| k 鈭� 鈩潁. (h) Eigenvalues: \(\lambda_1 = 1\), \(\lambda_2 = 3\), \(\lambda_3 = -1\). Eigenspaces: E(1) = {k\(\begin{bmatrix}1\\0\\0\end{bmatrix}\)| k 鈭� 鈩潁, E(3) = {k\(\begin{bmatrix}2\\1\\0\end{bmatrix}\)| k 鈭� 鈩潁, E(-1) = {k\(\begin{bmatrix}1\\-1\\5\end{bmatrix}\)| k 鈭� 鈩潁. (i) Eigenvalues: \(\lambda_1 = 2\), \(\lambda_2 = 2\), \(\lambda_3 = -1\). Eigenspaces: E(2) = {k\(\begin{bmatrix}5\\2\\1\end{bmatrix}\)| k 鈭� 鈩潁, E(-1) = {k\(\begin{bmatrix}-1\\1\\1\end{bmatrix}\)| k 鈭� 鈩潁. (j) Eigenvalues: \(\lambda_1 = 1\), \(\lambda_2 = -2\), \(\lambda_3 = -1\). Eigenspaces: E(1) = {k\(\begin{bmatrix}1\\0\\0\end{bmatrix}\)| k 鈭� 鈩潁, E(-2) = {k\(\begin{bmatrix}0\\1\\1\end{bmatrix}\)| k 鈭� 鈩潁, E(-1) = {k\(\begin{bmatrix}1\\1\\1\end{bmatrix}\)| k 鈭� 鈩潁. (k) Eigenvalues: \(\lambda_1 = 2\), \(\lambda_2 = 2\), \(\lambda_3 = 3\), \(\lambda_4 = 4\). Eigenspaces: E(2) = {k\(\begin{bmatrix}1\\0\\0\\0\end{bmatrix}\)| k 鈭� 鈩潁, E(3) = {k\(\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\)| k 鈭� 鈩潁, E(4) = {k\(\begin{bmatrix}0\\0\\0\\1\end{bmatrix}\)| k 鈭� 鈩潁. (l) Eigenvalues: \(\lambda_1 = 3\), \(\lambda_2 = 3\), \(\lambda_3 = 2\), \(\lambda_4 = 2\). Eigenspaces: E(3) = {k\(\begin{bmatrix}1\\0\\0\\0\end{bmatrix}\)| k 鈭� 鈩潁, E(2) = {k\(\begin{bmatrix}0\\0\\1\\0\end{bmatrix}\)| k 鈭� 鈩潁.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Eigenspaces
Eigenspaces are a crucial concept when dealing with eigenvectors and eigenvalues. An eigenspace is the set of all eigenvectors that correspond to a given eigenvalue, along with the zero vector. In essence, it is a subspace of a vector space. The importance of eigenspaces lies in their ability to simplify complex linear transformations and systems.

An eigenvector is considered a direction in which a transformation acts like a simple scaling, and all vectors that are multiples of this eigenvector form the eigenspace. For a given matrix \( A \) and an eigenvalue \( \lambda \), the eigenspace \( E(\lambda) \) can be found by solving the equation \((A - \lambda I)\mathbf{x} = 0\). This gives the set of all solutions \( \mathbf{x} \), known as the eigenspace for \( \lambda \).

The dimension of an eigenspace, also called the geometric multiplicity of an eigenvalue, provides insights into the structure and properties of the matrix. Understanding eigenspaces helps you grasp the behavior of linear transformations on different subspaces in a vector space.
Characteristic Polynomial
The characteristic polynomial is a central tool in finding eigenvalues of a matrix. It is derived from the matrix itself by constructing the polynomial equation \(|A - \lambda I|\), where \( A \) is the given matrix and \( I \) is the identity matrix of the same dimension.

The determinant of the matrix \( (A - \lambda I) \) gives us the characteristic polynomial, which is typically in a quadratic or higher-degree form, depending on the size of \( A \). This polynomial encapsulates information about the matrix's eigenvalues (the roots of the polynomial) which are crucial to understanding the matrix's behavior in various linear transformations.

Solving the characteristic polynomial equation allows us to find the eigenvalues, which are the critical first step in determining the eigenspaces and eigenvectors for the matrix. Studying the characteristic polynomial will enable you to predict how a matrix transforms a vector space.
Quadratic Formula
The quadratic formula is a reliable method for solving quadratic equations, and it's often used to find the eigenvalues from the characteristic polynomial. A typical quadratic equation is expressed in the standard form \( ax^2 + bx + c = 0 \) and the quadratic formula \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \) helps find the values of \( x \).

In the context of eigenvalues, once the characteristic polynomial has been reduced to a quadratic form, the quadratic formula provides a method to calculate the eigenvalues. These values are pivotal for further calculations, such as determining eigenspaces and eigenvectors.

It is important to note that the discriminant \( b^2 - 4ac \) in the quadratic formula indicates the nature of the eigenvalues:
  • If positive, there are two distinct real eigenvalues.
  • If zero, there are duplicate (repeated) real eigenvalues.
  • If negative, the eigenvalues are complex.
Understanding the quadratic formula and how it applies to characteristic polynomials is key to mastering the calculations of eigenvalues in linear algebra.
Eigenvectors
Eigenvectors, as solutions to the eigenvalue equation \((A - \lambda I)\mathbf{x} = 0\), represent directions in which a matrix operation acts as a simple scaling. To find an eigenvector for a given eigenvalue \( \lambda \), it involves solving this system of linear equations.

For each eigenvalue found from the characteristic polynomial, there are one or more eigenvectors depending on the geometric multiplicity of the eigenvalue. Eigenvectors are crucial because they allow us to understand how a matrix changes a vector, often simplifying the analysis of linear transformations.

Once an eigenvalue \( \lambda \) is known, substitute it back into the matrix equation to find its eigenvector. These vectors can then be used to form eigenbases, which significantly simplify matrix operations, especially in higher-dimensional spaces.

To verify whether a vector is truly an eigenvector, substitute it into the eigenvalue equation and see if it satisfies \((A - \lambda I)\mathbf{x} = 0\). This step ensures accuracy when determining eigenvectors and allows further exploration into the transformation characteristics of matrices.

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Most popular questions from this chapter

Use the definition of the matrix exponential to compute \(e^{A}\) for each of the following matrices: (a) \(A=\left(\begin{array}{rr}1 & 1 \\ -1 & -1\end{array}\right)\) (b) \(A=\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)\) (c) \(A=\left(\begin{array}{rrr}1 & 0 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right)\)

Let \(B\) be an \(m \times n\) matrix of rank \(n .\) Show that \(B^{T} B\) is positive definite.

Show that if \(A\) is symmetric positive definite, then \(\operatorname{det}(A)>0 .\) Give an example of a \(2 \times 2\) matrix with positive determinant that is not positive definite.

The city of Mawtookit maintains a constant population of 300,000 people from year to year. A political science study estimated that there were 150,000 Independents, 90,000 Democrats, and 60,000 Republicans in the town. It was also estimated that each year 20 percent of the Independents become Democrats and 10 percent become Republicans. Similarly, 20 percent of the Democrats become Independents and 10 percent become Republicans, while 10 percent of the Republicans defect to the Democrats and 10 percent become Independents each year. Let \\[ \mathbf{x}=\left(\begin{array}{r} 150,000 \\ 90,000 \\ 60,000 \end{array}\right) \\] and let \(\mathbf{x}^{(1)}\) be a vector representing the number of people in each group after one year (a) Find a matrix \(A\) such that \(A \mathbf{x}=\mathbf{x}^{(1)}\) (b) Show that \(\lambda_{1}=1.0, \lambda_{2}=0.5,\) and \(\lambda_{3}=0.7\) are the eigenvalues of \(A,\) and factor \(A\) into a product \(X D X^{-1},\) where \(D\) is diagonal (c) Which group will dominate in the long run? Justify your answer by computing \(\lim _{n \rightarrow \infty} A^{n} \mathbf{x}\)

Let \(A\) be an \(n \times n\) matrix with singular value decomposition \(U \Sigma V^{T}\) and let \\[ B=\left(\begin{array}{cc} O & A^{T} \\ A & O \end{array}\right) \\] Show that if \\[ \mathbf{x}_{i}=\left(\begin{array}{c} \mathbf{v}_{i} \\ \mathbf{u}_{i} \end{array}\right], \quad \mathbf{y}_{i}=\left[\begin{array}{r} -\mathbf{v}_{i} \\ \mathbf{u}_{i} \end{array}\right], \quad i=1, \ldots, n \\] then the \(\mathbf{x}_{i}\) 's and \(\mathbf{y}_{i}\) 's are eigenvectors of \(B\). How do the eigenvalues of \(B\) relate to the singular values of \(A ?\)
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