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Chapter 5: Problem 1

Consider the matrices $$ A_{1}=\left[\begin{array}{rrr} 1 & -1 & -i \\ -1 & 1 & -1 \\ -1 & -1 & 1 \end{array}\right], A_{2}=\left[\begin{array}{rrr} 3 & -2 & 0 \\ -2 & 3 & 0 \\ 0 & 0 & 5 \end{array}\right] $$ and the vectors $$ \begin{aligned} &\mathbf{v}_{1}=\left[\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right], \mathbf{v}_{2}=\left[\begin{array}{l} 0 \\ 0 \\ 1 \end{array}\right], \mathbf{v}_{3}=\left[\begin{array}{r} -1 \\ 1 \\ 0 \end{array}\right], \\ &\mathbf{v}_{4}=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right], \mathbf{v}_{5}=\left[\begin{array}{r} -1 \\ 0 \\ 1 \end{array}\right] . \end{aligned} $$ List the vectors that are eigenvectors of \(A_{1}\) and the ones that are eigenvectors of \(A_{2}\). Give the eigen value in each case.

Short Answer

Expert verified
\( \mathbf{v}_2 \) and \( \mathbf{v}_4 \) are eigenvectors of \( A_1 \) with eigenvalues \(-i\) and \(-2\) respectively.

Step by step solution

01

Calculate Eigenvectors for Matrix A1

To determine if a vector \( \mathbf{v} \) is an eigenvector of a matrix \( A_1 \), we check if \( A_1\mathbf{v} = \lambda \mathbf{v} \) for some scalar \( \lambda \). Let's test all given vectors against matrix \( A_1 \):1. **Test \( \mathbf{v}_1 \):** \[ A_1 \mathbf{v}_1 = \begin{bmatrix} 1 & -1 & -i \ -1 & 1 & -1 \ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} = \begin{bmatrix} 1-1-i \ -1+1-1 \ -1-1+1 \end{bmatrix} = \begin{bmatrix}-i \ -1 \ -1 \end{bmatrix} eq \lambda \begin{bmatrix} 1 \ 1 \ 1 \end{bmatrix} \] Therefore, \( \mathbf{v}_1 \) is not an eigenvector of \( A_1 \).2. **Test \( \mathbf{v}_2 \):** \[ A_1 \mathbf{v}_2 = \begin{bmatrix} 1 & -1 & -i \ -1 & 1 & -1 \ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 0 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} -i \ -1 \ 1 \end{bmatrix} \] This simplifies to \( A_1 \mathbf{v}_2 = -i \mathbf{v}_2 \) showing \( \lambda = -i\), thus \( \mathbf{v}_2 \) is an eigenvector with eigenvalue \( -i \).3. **Test \( \mathbf{v}_3 \):** \[ A_1 \mathbf{v}_3 = \begin{bmatrix} 1 & -1 & -i \ -1 & 1 & -1 \ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} -1 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} -1+1 \ 1-1 \ -1-1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ -2 \end{bmatrix} eq \lambda \begin{bmatrix} -1 \ 1 \ 0 \end{bmatrix} \] Therefore, \( \mathbf{v}_3 \) is not an eigenvector of \( A_1 \).4. **Test \( \mathbf{v}_4 \):** \[ A_1 \mathbf{v}_4 = \begin{bmatrix} 1 & -1 & -i \ -1 & 1 & -1 \ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} = \begin{bmatrix} 1-1 \ -1+1 \ -1-1 \end{bmatrix} = \begin{bmatrix} 0 \ 0 \ -2 \end{bmatrix} = -2\begin{bmatrix} 1 \ 1 \ 0 \end{bmatrix} \] Thus, \( \mathbf{v}_4 \) is an eigenvector with eigenvalue \( -2 \).5. **Test \( \mathbf{v}_5 \):** \[ A_1 \mathbf{v}_5 = \begin{bmatrix} 1 & -1 & -i \ -1 & 1 & -1 \ -1 & -1 & 1 \end{bmatrix} \begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix} = \begin{bmatrix} 1-i \ 1-1 \ 0 \end{bmatrix} eq \lambda \begin{bmatrix} -1 \ 0 \ 1 \end{bmatrix} \] Therefore, \( \mathbf{v}_5 \) is not an eigenvector of \( A_1 \).

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Key Concepts

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

Matrices
Matrices are rectangular arrays of numbers arranged in rows and columns typically used in linear algebra to represent linear transformations and solve systems of linear equations. A matrix is often denoted by uppercase letters such as \( A, B, \) etc. The elements within a matrix are defined by a holistic organization across its rows and columns. For instance, a matrix like\[A = \begin{bmatrix}a_{11} & a_{12} & \cdots & a_{1n} \a_{21} & a_{22} & \cdots & a_{2n} \\vdots & \vdots & \ddots & \vdots \a_{m1} & a_{m2} & \cdots & a_{mn}\end{bmatrix}\]Here, \( A \) has \( m \) rows and \( n \) columns, making it an \( m \times n \) matrix. In the context of eigenvectors and eigenvalues, matrices help in determining the relationship between different vector transformations.
Eigenvalues
Eigenvalues are scalars associated with a linear transformation represented by a matrix. They are fundamental in determining the behavior of matrices when applied to vector spaces.When a vector remains parallel to itself after a linear transformation, it implies the presence of an eigenvalue - a special scalar that describes the vector's stretch or compression.Consider a vector \( \mathbf{v} \) and a matrix \( A \).Finding the eigenvalue \( \lambda \) involves verifying the equation: \[A\mathbf{v} = \lambda \mathbf{v}\]. It shows that applying matrix \( A \) to vector \( \mathbf{v} \) scales it by \( \lambda \).
  • If \( \lambda = 1 \): \( \mathbf{v} \) retains its original length.
  • If \( \lambda > 1 \): \( \mathbf{v} \) is stretched.
  • If \( \lambda < 1 \): \( \mathbf{v} \) is compressed.
  • If \( \lambda = 0 \): \( \mathbf{v} \) becomes the zero vector.
These characteristics allow eigenvalues to play a pivotal role in analyzing stability, vibrations, and more.
Linear Algebra
Linear algebra is the branch of mathematics concerning vector spaces and linear mappings between them. Its core focuses on understanding concepts like vectors, vector spaces, and matrices to solve systems of linear equations efficiently. Besides providing techniques for solving equations, it extends its application to various fields like physics, computer science, and engineering.
In linear algebra, the significance of matrices and eigen structures cannot be overemphasized. Matrices translate linear transformations, making them tools for encoding geometrical and algebraic changes. Understanding the concept of eigenvectors and eigenvalues allows analysts to comprehend how these transformations impact different vector spaces.
Vector Spaces
A vector space is a fundamental concept in linear algebra that refers to a collection of vectors. These vectors are closed under the operations of addition and scalar multiplication. They satisfy certain properties like commutativity, associativity, and distributivity.
In a vector space, given any two vectors \( \mathbf{u} \) and \( \mathbf{v} \), and any scalar \( c \), the following properties apply:
  • \( \mathbf{u} + \mathbf{v} \) is still a vector in the same space.
  • \( c\mathbf{u} \) remains a vector within the vector space.
  • The zero vector serves as an identity element, meaning adding it to any vector doesn鈥檛 change the vector.
  • Each vector has an additive inverse, allowing subtraction between vectors in the same space.
Understanding vector spaces is essential when analyzing how different transformations, like those represented by matrices, affect a set of vectors.

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Most popular questions from this chapter

Solve the given system of linear differential equations as outlined in the summary: $$ \begin{aligned} &x_{1}^{\prime}=x_{1}+4 x_{2} \\ &x_{2}^{\prime}=3 x_{1} \end{aligned} $$

Mark each of the following True or False. \- a. 脡very \(n \times n\) matrix is diagonalizabie. If an \(n \times n\) matrix has \(n\) distinct real eigenvalues, it is diagonalizable. c. Every \(n \times n\) real symmetric matrix is teal diagonalizable. d. An \(n \times n\) matrix is diagonalizable if and only if it has \(n\) distinct eigenvalues. i. If an \(n \times n\) matrix \(A\) is diagonalizable, there is a unique diagonal matrix \(D\) that is similar to \(A\). If \(A\) and \(B\) are similar square matrices, then \(\operatorname{det}(A)=\operatorname{det}(B)\). \- e. An \(n \times n\) matrix is diagonalizable if and only if the algebraic multiplicity of each of its eigenvalues equals the geometric multiplicity. -f. Every invertible matrix is diagonalizabie. \- g. Every triangular matrix is diagonalizable. h. If \(A\) and \(B\) are similar square matrices and \(A\) is diagonalizable, then \(B\) is also diagonalizable.

The analogue of Exercise 41 for a list of \(r\) eigenvectors in \(V\) having distinct eigenvalues is also true; that is, the vectors are independent. See if you can prove it. [Hint: Suppose that the vectors are dependent; consider the first vector in the list that is a linear combination of its predecessors, and apply \(T\).]

In Exercises 2-16, find the characteristic polynomial, the real eigenvalues, and the corresponding eigenvectors of the given matrix. $$ \left[\begin{array}{rrr} -1 & 0 & 0 \\ -4 & 2 & -1 \\ 4 & 0 & 3 \end{array}\right] $$

Mark each of the following True or False. a. Every square matrix has real eigenvalues. b. Every \(n \times n\) matrix has \(n\) distinct (possibly complex) eigenvalues. c. Every \(n \times n\) matrix has \(n\) not necessarily distinct and possibly complex eigenvalues. d. There can be only one eigenvalue associated with an eigenvector of a linear transformation. e. There can be only one eigenvector associated with an eigenvalue of a linear transformation. f. If \(\mathbf{v}\) is an eigenvector of a matrix \(A\), then \(\mathbf{v}\) is an eigenvector of \(A+c I\) for all scalars \(c\). g. If \(\lambda\) is an eigenvalue of a matrix \(A\), then \(\lambda\) is an eigenvalue of \(A+c I\) for all scalars \(c\). h. If \(\mathbf{v}\) is an eigenvector of an invertible matrix \(A\), then \(c v\) is an eigenvector of \(A^{-1}\) for all nonzero scalars \(c\). i. Every vector in a vector space \(V\) is an eigenvector of the identity transformation of \(V\) into \(V\). j. Every nonzero vector in a vector space \(V\) is an eigenvector of the identity transformation of \(V\) into \(V\).
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