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Chapter 7: Problem 37

Consider an eigenvalue \(\lambda_{0}\) of an \(n \times n\) matrix \(A .\) We are told that the algebraic multiplicity of \(\lambda_{0}\) exceeds 1. Show that \(f_{A}^{\prime}\left(\lambda_{0}\right)=0\) (i.e., the derivative of the characteristic polynomial of \(A\) vanishes at \(\lambda_{0}\) ).

Short Answer

Expert verified
Since \(\lambda_0\) has an algebraic multiplicity greater than 1, the characteristic polynomial includes the factor \( (\lambda - \lambda_0)^2\) or higher. The derivative of this polynomial necessarily includes \( (\lambda - \lambda_0)\) as a factor. Therefore, evaluating the derivative at \(\lambda_0\) gives 0, showing \(f_A'(\lambda_0) = 0\).

Step by step solution

01

Define the characteristic polynomial

Define the characteristic polynomial of the matrix A, denoted as \(f_A(\text{\textbackslash}lambda)\). The characteristic polynomial is obtained by taking the determinant of the matrix \((A - \text{\textbackslash}lambda I)\), where \(I\) is the identity matrix of the same size as \(A\). Thus, \(f_A(\text{\textbackslash}lambda) = \text{\textbackslash}det(A - \text{\textbackslash}lambda I) = 0\).
02

State the algebraic multiplicity condition

Since we are given that the algebraic multiplicity of the eigenvalue \(\text{\textbackslash}lambda_0\) exceeds 1, this means that \((\text{\textbackslash}lambda - \text{\textbackslash}lambda_0)\) is a factor of the characteristic polynomial \(f_A(\text{\textbackslash}lambda)\) and it is present at least twice, i.e., \((\text{\textbackslash}lambda - \text{\textbackslash}lambda_0)^2\) divides \(f_A(\text{\textbackslash}lambda)\).
03

Differentiate the characteristic polynomial

Differentiate the characteristic polynomial \(f_A(\text{\textbackslash}lambda)\). Given that \(f_A(\text{\textbackslash}lambda)\) has \((\text{\textbackslash}lambda - \text{\textbackslash}lambda_0)^2\) as a factor, after applying the product rule while differentiating, there will always be at least one \((\text{\textbackslash}lambda - \text{\textbackslash}lambda_0)\) term present in the derivative.
04

Evaluate the derivative at \(\text{\textbackslash}lambda_0\)

Evaluate \(f_A'(\text{\textbackslash}lambda)\) at \(\text{\textbackslash}lambda_0\). As the derivative \(f_A'(\text{\textbackslash}lambda)\) contains at least one factor of \((\text{\textbackslash}lambda - \text{\textbackslash}lambda_0)\), substituting \(\text{\textbackslash}lambda = \text{\textbackslash}lambda_0\) into the derivative will result in \(f_A'(\text{\textbackslash}lambda_0) = 0\) because \((\text{\textbackslash}lambda_0 - \text{\textbackslash}lambda_0) = 0\). This shows that the derivative of the characteristic polynomial vanishes at the eigenvalue \(\text{\textbackslash}lambda_0\).

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Key Concepts

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

Eigenvalues and Eigenvectors
When it comes to understanding the behavior of linear transformations represented by matrices, eigenvalues and eigenvectors are fundamental concepts.

An eigenvalue is a scalar quantity \( \lambda \) that satisfies the equation \( A\vec{v} = \lambda\vec{v} \) where \( A \) is a square matrix, and \( \vec{v} \) is the corresponding eigenvector. The eigenvector \( \vec{v} \) is a non-zero vector that gets scaled by the factor of the eigenvalue when the matrix \( A \) is applied to it. Essentially, the action of the matrix on the eigenvector \( \vec{v} \) is the same as scaling \( \vec{v} \) by the eigenvalue.

It's crucial for students to recognize that finding eigenvalues involves solving the characteristic polynomial for \( \lambda \). Each solution for this polynomial is an eigenvalue of the matrix. Jointly, an eigenvalue together with its eigenvector provides insights into the matrix's properties, such as its orientation, scaling, and whether it can be inverted.
Algebraic Multiplicity
In the context of eigenvalues, the term 'algebraic multiplicity' refers to the number of times an eigenvalue appears as a root of the matrix's characteristic polynomial. For example, if the characteristic polynomial is \( f_A(\lambda) = (\lambda - \lambda_0)^k \) for some eigenvalue \( \lambda_0 \) and integer \( k \) higher than 1, then \( \lambda_0 \) is said to have an algebraic multiplicity of \( k \).

Understanding the concept of algebraic multiplicity is important because it relates to the dimensionalities of eigenspaces. It can be indicative of the matrix's sensitivity to perturbations and the independence of its eigenvectors. A higher algebraic multiplicity suggests a richer geometric multiplicity but this is not always the case, and it's also an area which often trips up students. Recognizing the difference between algebraic and geometric multiplicity and how they relate to each other is essential to mastery of linear algebra.
Determinant of a Matrix
The determinant is a scalar value that is a function of the entries of a square matrix. It can be considered a scaling factor for the transformation that the matrix represents, and it carries a wealth of information about the matrix.

For example, a determinant of zero indicates that the matrix does not have an inverse, and thus is not full rank, signifying a loss of dimensionality in the transformation. Conversely, a non-zero determinant implies that the matrix is invertible.

When we refer to the characteristic polynomial in the context of eigenvalues, we actually speak about the determinant of a shifted matrix, \( A - \lambda I \) where \( I \) is the identity matrix. Finding where this determinant equals zero provides the eigenvalues of the matrix \( A \) because it's at these points that the matrix \( A - \lambda I \) becomes singular and ceases to be invertible. Understanding how the determinant relates to the properties of a matrix, such as invertibility and eigenvalues, is an important skill in linear algebra.

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Most popular questions from this chapter

Two interacting populations of hares and foxes can be modeled by the recursive equations $$\begin{array}{l} h(t+1)=4 h(t)-2 f(t) \\ f(t+1)=h(t)+f(t) \end{array}$$ For each of the initial populations given in parts (a) through (c), find closed formulas for \(h(t)\) and \(f(t)\) a. \(h(0)=f(0)=100\) b. \(h(0)=200, f(0)=100\) c. \(h(0)=600, f(0)=500\)

Consider a \(5 \times 5\) matrix \(A\) and a vector \(\vec{v}\) in \(\mathbb{R}^{5}\). Suppose the vectors \(\vec{v}, A \vec{v}, A^{2} \vec{v}\) are linearly independent, while \(A^{3} \vec{v}=a \vec{v}+b A \vec{v}+c A^{2} \vec{v}\) for some scalars \(a, b, c . \mathrm{We}\) can take the linearly independent vectors \(\vec{v}, A \vec{v}, A^{2} \vec{v}\) and expand them to a basis \(\mathfrak{B}=\left(\vec{v}, A \vec{v}, A^{2} \vec{v}, \vec{w}_{4}, \vec{w}_{5}\right)\) of \(\mathbb{R}^{5}\). a. Consider the matrix \(B\) of the linear transformation \(T(\vec{x})=A \vec{x}\) with respect to the basis \(\mathfrak{B}\) Write the entries of the first three columns of \(B\). (Note that we do not know anything about the entries of the last two columns of \(B .)\) b. Explain why \(f_{A}(\lambda)=f_{B}(\lambda)=h(\lambda)\left(-\lambda^{3}+c \lambda^{2}+\right.\) \(b \lambda+a),\) for some quadratic polynomial \(h(\lambda) .\) See Exercise 51. c. Explain why \(f_{A}(A) \vec{v}=\overrightarrow{0} .\) Here, \(f_{A}(A)\) is the characteristic polynomial evaluated at \(A\), that is, if \(f_{A}(\lambda)=c_{n} \lambda^{n}+\cdots+c_{1} \lambda+c_{0},\) then \(f_{A}(A)=\) \(c_{n} A^{n}+\cdots+c_{1} A+c_{0} I_{n}\).

Consider the dynamical system $$\vec{x}(t+1)=\left[\begin{array}{cc} 1.1 & 0 \\ 0 & \lambda \end{array}\right] \vec{x}(t)$$ Sketch a phase portrait of this system for the given values of \(\lambda:\) $$\lambda=1$$

Find all complex eigenvalues of the matrices in Exercises 20 through 26 (including the real ones, of course). Do not use technology. Show all your work. $$\left[\begin{array}{ll} 3 & -5 \\ 2 & -3 \end{array}\right]$$

Consider the set \(\mathbb{H}\) of all complex \(2 \times 2\) matrices of the form \\[ A=\left[\begin{array}{rr} w & -\bar{z} \\ z & \bar{w} \end{array}\right] \\] where \(w\) and \(z\) are arbitrary complex numbers. a. Show that \(\mathbb{H}\) is closed under addition and multiplication. (That is, show that the sum and the product of two matrices in \(\mathbb{H}\) are again in \(\mathbb{H}\). .) b. Which matrices in \(\mathbb{H}\) are invertible? c. If a matrix in \(\mathbb{H}\) is invertible, is the inverse in \(\mathbb{H}\) as well? d. Find two matrices \(A\) and \(B\) in \(\mathbb{H}\) such that \(A B \neq\) \(B A\) \(\mathbb{H}\) is an example of a skew field: It satisfies all axioms for a field, except for the commutativity of multiplication. [The skew field \(\mathbb{H}\) was introduced by the Irish mathematician Sir William Hamilton \((1805-1865) ;\) its elements are called the quaternions. Another way to define the quaternions is discussed in Exercise \(5.3 .64 .\)
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