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Chapter 7: Problem 55

Prove that the constant term of the characteristic polynomial is \(\pm|A|\).

Short Answer

Expert verified
From the definition of the characteristic polynomial and its standard form, it is clear that setting \(\lambda = 0\) gives the constant term, which is equivalent to the determinant of the matrix \(A\). This term can be positive or negative based on the order of the matrix.

Step by step solution

01

Write Down the Definition of The Characteristic Polynomial

The characteristic polynomial \(p(\lambda)\) of a given square matrix \(A\) is defined by the determinant of the matrix \(A-\lambda I\), where \(I\) is the identity matrix. That is, \(p(\lambda) = |A-\lambda I|\).
02

Express the Characteristic Polynomial in Standard Form

The characteristic polynomial is a polynomial of degree \(n\) (where \(n\) is the order of matrix \(A\)), which can be expressed in the standard form as \(p(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + a_{n-2}\lambda^{n-2} + ... + a_1\lambda + a_0\). Here, \(a_0\) is the constant term.
03

Identify the Constant Term

From the standard form, it's clear that the constant term is the result of setting \(\lambda = 0\) in the characteristic polynomial. So, \(a_0 = p(0) = |A-0I| = |A|\).
04

Mention the Effect of the Matrix Order on the Sign of the Constant Term

Notice that since the characteristic polynomial is of degree \(n\), when the order of the matrix \(n\) is odd, the constant term will take a negative sign due to the \(-\lambda\) term in the determinant. Thus, the constant term of the characteristic polynomial is \(\pm|A|\).

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Most popular questions from this chapter

Write out the system of first-order linear differential equations represented by the matrix equation \(\mathbf{y}^{\prime}=A \mathbf{y} .\) Then verify the indicated general solution. $$A=\left[\begin{array}{rrr} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & -4 & 0 \end{array}\right], \begin{array}{rr} y_{1}=C_{1}+ & C_{2} \cos 2 t+C_{3} \sin 2 t \\ y_{2}= & 2 C_{3} \cos 2 t-2 C_{2} \sin 2 t \\ y_{3}= & -4 C_{2} \cos 2 t-4 C_{3} \sin 2 t \end{array}$$

Solve the system of first-order linear differential equations. $$\begin{array}{l} y_{1}^{\prime}=y_{1}+2 y_{2} \\ y_{2}^{\prime}=2 y_{1}+y_{2} \end{array}$$

Prove that if a symmetric matrix \(A\) has only one eigenvalue \(\lambda\) then \(A=\lambda I\).

Solve the system of first-order linear differential equations. $$\begin{array}{l} y_{1}^{\prime}=-y_{1} \\ y_{2}^{\prime}=6 y_{2} \end{array}$$

Find an orthogonal matrix \(P\) such that \(P^{T} A P\) diagonalizes \(A .\) Verify that \(P^{T} A P\) gives the proper diagonal form. $$A=\left[\begin{array}{lll} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{array}\right]$$
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