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Chapter 11: Problem 22

Show that each polynomial \(a_{0}-a_{1} \lambda+a_{2} \lambda^{2}-\ldots+(-1)^{\mathrm{n}} \lambda^{\mathrm{n}}(n \geqslant 1)\) is the characteristic polynomial of some suitable matrix. [Hint: what is the characteristic polynomial of $$ \left.\left[\begin{array}{lll} 0 & 1 & 0 \\ 0 & 0 & 1 \\ c & b & a \end{array}\right] ?\right] $$

Short Answer

Expert verified
Each polynomial is the characteristic polynomial of its corresponding companion matrix.

Step by step solution

01

Understand the Matrix Structure

We start by examining the structure of the matrix provided in the hint, which is a 3x3 matrix:\[ A = \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ c & b & a \end{bmatrix} \]This matrix has a specific pattern where the subdiagonal entries are zeros, and there is one non-zero element on each column moving down one row, while the last row has variables \(c, b, a\).
02

Find the Characteristic Polynomial

The characteristic polynomial is found by calculating the determinant of \(A - \lambda I\), where \(I\) is the identity matrix of the same size. Therefore, the matrix \(A - \lambda I\) is:\[ A - \lambda I = \begin{bmatrix} -\lambda & 1 & 0 \ 0 & -\lambda & 1 \ c & b & a - \lambda \end{bmatrix} \]To find the characteristic polynomial, compute the determinant of this 3x3 matrix.
03

Calculate the Determinant

The determinant of a 3x3 matrix \(\begin{bmatrix} e & f & g \ h & i & j \ k & l & m \end{bmatrix}\) is computed as:\[e(i m - j l) - f(h m - j k) + g(h l - i k)\]Apply this to the matrix \(A - \lambda I\):- First element \(-\lambda\) multiplies the determinant of the lower right 2x2 matrix, which is \((-\lambda)(a - \lambda) - b\cdot 1\).- Subtract the second element \(1\) times the determinant of the 2x2 matrix excluding the first row, second column, which is \(0\cdot (a - \lambda) - (c)(1)\).- The third term involves \(0\) which gives no contribution.The computation simplifies to finding:\[-\lambda((-\lambda)(a - \lambda) - b) + c\]
04

Simplify the Polynomial Expression

Perform the simplification:\[-\lambda((\lambda^2 - a\lambda) - b) + c = \lambda^3 - a\lambda^2 + b\lambda + c\]Thus, the given polynomial is \(\lambda^3 - a\lambda^2 + b\lambda + c\), identical to the polynomial pattern given: \(a_0 - a_1 \lambda + a_2 \lambda^2 - \ldots + (-1)^n \lambda^n\).
05

Generalize for Higher Order Polynomials

For a polynomial of degree \(n\), consider an \(n\times n\) companion matrix, which has the form:\[ \begin{bmatrix} 0 & 1 & 0 & \ldots & 0 \ 0 & 0 & 1 & \ldots & 0 \ \vdots & \vdots & \vdots & \ddots & \vdots \ 0 & 0 & 0 & \ldots & 1 \ c_n & c_{n-1} & c_{n-2} & \ldots & c_1 \end{bmatrix} \]The characteristic polynomial of this matrix is exactly the polynomial given in the question.

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Key Concepts

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

Companion Matrix
The concept of a companion matrix is central in linear algebra, particularly when dealing with polynomials. It's a specific type of square matrix formed based on the coefficients of a polynomial. The role of the companion matrix is to facilitate the determination of the characteristic polynomial associated with a matrix. In our exercise, the given form of the companion matrix is: \[ \begin{bmatrix} 0 & 1 & 0 \ 0 & 0 & 1 \ c & b & a \end{bmatrix} \] This matrix is constructed such that its characteristic polynomial matches a given polynomial. The last row contains the coefficients of the polynomial in reverse order, except for the constant term which aligns with the higher-degree coefficient. Everywhere else, the matrix is filled with a pattern of 0's and 1's in a diagonal line, stepping through the middle of the matrix. Companion matrices are incredibly useful for understanding the roots and behavior of polynomials through matrix operations.
Determinant Calculation
Calculating the determinant of a matrix is essential for finding characteristic polynomials. The determinant provides valuable insights into the properties of a matrix, such as singularity and invertibility. For a 3x3 matrix like the one in our exercise, the determinant is computed using the formula: \[e(i m - j l) - f(h m - j k) + g(h l - i k)\] This method involves breaking down the matrix into smaller 2x2 matrices and calculating minors and cofactors. In the example, the process was used to find the determinant of \(A - \lambda I\), where the computed result gave the characteristic polynomial. By executing determinant calculations, we derive the polynomial equation that features the eigenvalues of the matrix as its roots.
Matrix Polynomial
Matrix polynomials extend the idea of polynomials to matrices, allowing operations to be performed that are similar to those on numbers. A matrix polynomial is expressed in terms of powers of a matrix, combined using scalar coefficients. In our solution, the constructed polynomial \( \lambda^3 - a\lambda^2 + b\lambda + c \) derives directly from the structure of the matrix.
  • \( \lambda^3 \) corresponds to \( \lambda I \), the identity matrix scaled by \( \lambda^3 \).
  • \( -a\lambda^2 \) comes from the product of \( -a \) and \( \lambda^2 \).
  • \( +b\lambda \) assigns the second coefficient.
  • \( +c \) aligns the constant term.
These steps ensure that the polynomial captures the essential characteristics of the matrix, facilitating the diagnosis of matrix properties through polynomial analysis.
Linear Algebra Problem-Solving
Linear algebra involves various techniques and tools for solving complex problems, especially those related to matrices and polynomials. This exercise exemplifies practical problem-solving techniques by applying abstract algebraic concepts to very concrete computations.
  • Understand the Problem: Start by interpreting the given problem and identifying applicable concepts, such as matrix structures and polynomial forms.
  • Utilize Matrix Properties: Leverage properties such as determinants and eigenvalues to solve characteristic equations.
  • Generalize the Solution: Extend solutions from specific examples (like 3x3 matrices) to generic forms (like \(n\times n\) matrices), enabling solutions to a broader class of problems.
  • Simplify Expressions: Break down complex polynomial expressions into manageable components for easier computation and insight.
By systematically tackling each element of a problem, students can develop a comprehensive understanding of linear algebra concepts, paving the way for solving similar exercises with confidence and precision.

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Most popular questions from this chapter

When the Laboratory Party came to power \(42 \%\) of the population were 'poor', \(36 \%\) were 'middle income' and \(22 \%\) classified as 'rich'. The party claimed the number of poor would decrease if all future governments followed its policies based on the 5 -year transition matrix \(\begin{array}{cccc}\mathrm{p} & \mathrm{m} & \mathrm{r} & \\ 0.8 & 0.2 & 0.1 & \mathrm{p} \\ 0.1 & 0.7 & 0.2 & \mathrm{~m} \\ 0.1 & 0.1 & 0.7 & \mathrm{r}\end{array}\) which indicates, for example, of the p(oor), \(\frac{1}{10}\) will, in 5 years, become m(iddle income) and \(\frac{1}{10}\) become r(ich). Is this transition matrix compatible with the party's claims?

(i) Let \(A\) and \(B\) be \(n \times n\) matrices. Give examples to show that if \(\lambda, \mu\) are eigenvalues of \(A\) and \(B\) respectively then \(\lambda+\mu\) need not be an eigenvalue of \(A+B\) and \(\lambda \mu\) need not be an eigenvalue of \(A B\) (ii) Show that, if \(\mathbf{x}\) is an eigenvector of both \(A\) and \(B\), then \(\mathrm{x}\) is an eigenvector of \(A B\). (iii) If \(A\) and \(B\) are row equivalent do \(A, B\) necessarily have the same eigenvalues or eigenvectors?

(a) Check that \(\left[\begin{array}{l}1 \\ 3\end{array}\right]\) is an eigenvector for the matrix \(\left[\begin{array}{cc}8 & 9 \\ 12 & 31\end{array}\right] .\) What is the corresponding eigenvalue? (b) Check that \(\left[\begin{array}{c}-2 \\\ 1\end{array}\right],\left[\begin{array}{c}10 \\ -5\end{array}\right]\) and \(\left[\begin{array}{c}7 \\ -4\end{array}\right]\) are eigenvectors for \(\left[\begin{array}{cc}-9 & -28 \\ 8 & 21\end{array}\right] .\) Is \(\left[\begin{array}{c}-2 \\ 1\end{array}\right]+\left[\begin{array}{c}10 \\\ -5\end{array}\right]\) also an eigenvector for \(\left[\begin{array}{cc}-9 & -28 \\\ 8 & 21\end{array}\right] ?\) What about \(\left[\begin{array}{c}-2 \\\ 1\end{array}\right]+\left[\begin{array}{c}7 \\ -4\end{array}\right] ?\)

Let \(A\) be a diagonalisable matrix so that \(C^{-1} A C=D\) for suitable \(C\) and diagonal \(D\). Show that for each such \(D\) there are infinitely many such \(C\).

Show that the eigenvalues of a skew symmetric matrix are pure imaginary. (That is, are of the form \(a+i b\) with \(a=0\).)
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