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Chapter 9: Problem 20

Prove Theorem 9.6: The following are equivalent: (i) The scalar \(\lambda\) is an eigenvalue of \(A\) (ii) The matrix \(\lambda I-A\) is singular. (iii) The scalar \(\lambda\) is a root of the characteristic polynomial \(\Delta(t)\) of \(A\)

Short Answer

Expert verified
To prove Theorem 9.6, we showed that (i), (ii), and (iii) are equivalent. We demonstrated that if 位 is an eigenvalue of A (i), then the matrix 位I - A is singular (ii). If 位I - A is singular (ii), then 位 is a root of the characteristic polynomial 螖(t) of A (iii). And if 位 is a root of the characteristic polynomial (iii), then 位 is an eigenvalue of A (i). As we have proved all three implications, the statements (i), (ii), and (iii) are equivalent, completing the proof of Theorem 9.6.

Step by step solution

01

(i) implies (ii)

Suppose that 饾渾 is an eigenvalue of A, then there exists a non-zero vector x, such that Ax = 饾渾x. To show that 饾渾I - A is singular, we need to prove that det(位I - A) = 0. We can write the equation Ax = 饾渾x as: A x = 饾渾 I x, which can be rearranged as: (饾渾I - A) x = 0 Since x is a non-zero vector, it implies that the matrix 饾渾I - A has a non-trivial nullspace, and thus it is singular, which means det(饾渾I - A) = 0.
02

(ii) implies (iii)

If the matrix 饾渾I - A is singular, then its determinant is zero: det(饾渾I - A) = 0. The determinant of 饾渾I - A is equal to the characteristic polynomial at 饾渾, 螖(饾渾): 螖(饾渾) = det(饾渾I - A) Since det(饾渾I - A) = 0, it means that 螖(饾渾) = 0, so 饾渾 is a root of the characteristic polynomial.
03

(iii) implies (i)

If 饾渾 is a root of the characteristic polynomial 螖(t), then 螖(饾渾) = 0. Substituting the definition of the characteristic polynomial, we get: 螖(饾渾) = det(饾渾I - A) = 0 Since det(饾渾I - A) = 0, it means that the matrix 饾渾I - A is singular. As we have shown before, when (饾渾I - A) is singular, there exists a non-zero vector x, such that: (饾渾I - A) x = 0 Rearranging it back to the form Ax = 饾渾x shows that 饾渾 is an eigenvalue of A. Now we have proved that (i) implies (ii), (ii) implies (iii), and (iii) implies (i). Therefore, the three statements (i), (ii), and (iii) are equivalent, which completes the proof of Theorem 9.6.

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Key Concepts

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

Singular Matrices
A matrix is termed singular when it does not have an inverse. In simpler terms, you cannot "undo" its effects with another matrix. This condition happens when the determinant of the matrix is zero. The determinant is a special number calculated from a square matrix, and if it's zero, this means there's no unique solution to the matrix equation.

In the context of eigenvalues, the condition \[ ( \lambda I - A ) x = 0 \] indicates a singular matrix because there is a non-zero solution for \( x \). This tells us that the matrix \( \lambda I - A \) alters the vector \( x \) into a zero vector, which suggests the matrix has collapsed into a lower-dimensional form.

When dealing with eigenvalues, a singular matrix helps prove equivalence statements, indicating how eigenvalues are key to understanding the behavior of complex systems. The detection of singular matrices aids in confirming a particular scalar \( \lambda \) is an eigenvalue of matrix \( A \).
Characteristic Polynomial
The characteristic polynomial is a significant concept in linear algebra. It's a polynomial that is derived from a matrix and its roots are the eigenvalues of the matrix. To find the characteristic polynomial of a matrix \( A \), the polynomial is expressed as:\[\Delta(t) = \mathrm{det}(tI - A)\]Here, \( I \) is the identity matrix of the same size as \( A \). Solving this polynomial equation lets us find the values (roots) where the determinant becomes zero鈥攖hese roots are the eigenvalues of \( A \).

In simpler terms, the characteristic polynomial acts like a gatekeeper. It tells us which scalar values \( \lambda \) cause the matrix \( \lambda I - A \) to become singular. This is crucial for understanding the behavior of the solutions to matrix equations, as it directly impacts the eigenvalues, which in turn tell us valuable information about transformations and systems. By determining the eigenvalues, we can decipher matrix behaviors such as rotations and scaling transformations in different dimensions.
Linear Algebra Theorems
Linear algebra theorems provide a foundation for understanding complex mathematical behavior and solving matrix equations. These theorems often show equivalence relationships between various conditions. In the proof of Theorem 9.6, we see the beauty of linear algebra through the equivalence of the statements: a scalar \( \lambda \) being an eigenvalue, the matrix \( \lambda I - A \) being singular, and \( \lambda \) as a root of the characteristic polynomial.

This theorem emphasizes the interconnectedness of concepts in linear algebra. It demonstrates how eigenvalues provide insights not only into the matrix's behavior but also confirm certain conditions about the matrix itself. The reliance on determinates and singularity introduces a powerful lens to analyze matrices and their transformations.

Understanding and applying these kind of theorems allow students and researchers to bridge the gap between abstract mathematical theory and practical application. It equips them with tools to predict outcomes, analyze stability, and optimize systems based on matrix behavior.

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Most popular questions from this chapter

For each of the following quadratic forms \(q(x, y, z)\), find an orthogonal substitution expressing \(x, y, z\) in terms of variables \(r, s, t\), and find \(q(r, s, t)\) (a) \(q(x, y, z)=5 x^{2}+3 y^{2}+12 x z\) (b) \(q(x, y, z)=3 x^{2}-4 x y+6 y^{2}+2 x z-4 y z+3 z^{2}\)

For each of the following symmetric matrices \(B\), find its eigenvalues, a maximal orthogonal set \(S\) of eigenvectors, and an orthogonal matrix \(P\) such that \(D=P^{-1} B P\) is diagonal: (a) \(B=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right]\), (b) \(\quad B=\left[\begin{array}{rrr}2 & 2 & 4 \\ 2 & 5 & 8 \\ 4 & 8 & 17\end{array}\right]\)

Let \(A=\left[\begin{array}{ll}2 & 2 \\ 1 & 3\end{array}\right]\) (a) Find all eigenvalues and corresponding eigenvectors (b) Find a nonsingular matrix \(P\) such that \(D=P^{-1} A P\) is diagonal, and \(P^{-1}\) (c) Find \(A^{6}\) and \(f(A),\) where \(t^{4}-3 t^{3}-6 t^{2}+7 t+3\) (d) Find a "real cube root" of \(B-\) that is, a matrix \(B\) such that \(B^{3}=A\) and \(B\) has real eigenvalues.

Prove Theorem 9.1: Let \(f\) and \(g\) be polynomials. For any square matrix \(A\) and scalar \(k\), (i) \((f+g)(A)=f(A)+g(A)\), (iii) \((k f)(A)=k f(A)\), (ii) \((f g)(A)=f(A) g(A)\), (iv) \(f(A) g(A)=g(A) f(A)\) Suppose \(f=a_{n} t^{n}+\cdots+a_{1} t+a_{0}\) and \(g=b_{m} t^{m}+\cdots+b_{1} t+b_{0} .\) Then, by definition, $$ f(A)=a_{n} A^{n}+\cdots+a_{1} A+a_{0} I \quad \text { and } \quad g(A)=b_{m} A^{m 1}+\cdots+b_{1} A+b_{0} I $$ (i) Suppose \(m \leq n\) and let \(b_{i}=0\) if \(i>m\). Then $$ f+g=\left(a_{n}+b_{n}\right) t^{n}+\cdots+\left(a_{1}+b_{1}\right) t+\left(a_{0}+b_{0}\right) $$ Hence, $$ \begin{aligned} (f+g)(A) &=\left(a_{n}+b_{n}\right) A^{n}+\cdots+\left(a_{1}+b_{1}\right) A+\left(a_{0}+b_{0}\right) I \\ &=a_{n} A^{n}+b_{n} A^{n}+\cdots+a_{1} A+b_{1} A+a_{0} I+b_{0} I=f(A)+g(A) \end{aligned} $$ (ii) By definition, \(f g=c_{n+m} t^{n+m}+\cdots+c_{1} t+c_{0}=\sum_{k=0}^{n+m} c_{k} t^{k}\), where $$ c_{k}=a_{0} b_{k}+a_{1} b_{k-1}+\cdots+a_{k} b_{0}=\sum_{i=0}^{k} a_{i} b_{k-i} $$ Hence, \((f g)(A)=\sum_{k=0}^{n+m} c_{k} A^{k}\) and $$ f(A) g(A)=\left(\sum_{i=0}^{n} a_{i} A^{i}\right)\left(\sum_{j=0}^{m} b_{j} A^{j}\right)=\sum_{i=0}^{n} \sum_{j=0}^{m} a_{i} b_{j} A^{i+j}=\sum_{k=0}^{n+m} c_{k} A^{k}=(f g)(A) $$ (iii) By definition, \(k f=k a_{n} t^{n}+\cdots+k a_{1} t+k a_{0}\), and so $$ (k f)(A)=k a_{n} A^{n}+\cdots+k a_{1} A+k a_{0} I=k\left(a_{n} A^{n}+\cdots+a_{1} A+a_{0} I\right)=k f(A) $$ (iv) By (ii), \(g(A) f(A)=(g f)(A)=\left(f_{B}\right)(A)=f(A) g(A)\)

For each of the following symmetric matrices \(A\), find an orthogonal matrix \(P\) and a diagonal matrix \(D\) such that \(D=P^{-1} A P\) : (a) \(A=\left[\begin{array}{rr}5 & 4 \\ 4 & -1\end{array}\right]\) (b) \(A=\left[\begin{array}{rr}4 & -1 \\ -1 & 4\end{array}\right]\), (c) \(A=\left[\begin{array}{rr}7 & 3 \\ 3 & -1\end{array}\right]\)
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