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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

Can a matrix be similar to two different diagonal matrices? Explain your answer.

Find (a) the characteristic equation and (b) the eigenvalues (and corresponding eigenvectors) of the matrix. $$\left[\begin{array}{lll} 3 & 2 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{array}\right]$$

Find an orthogonal matrix \(P\) such that \(P^{T} A P\) diagonalizes \(A .\) Verify that \(P^{T} A P\) gives the proper diagonal form. $$A=\left[\begin{array}{rr} 2 & \sqrt{2} \\ \sqrt{2} & 1 \end{array}\right]$$

Demonstrate the Cayley-Hamilton Theorem for the given matrix. The Cayley- Hamilton Theorem states that a matrix satisfies its characteristic equation. For example, the characteristic equation of \(A=\left[\begin{array}{rr}1 & -3 \\\ 2 & 5\end{array}\right]\) is \(\lambda^{2}-6 \lambda+11=0,\) and by the theorem you have \(A^{2}-6 A+11 I_{2}=O\). $$\left[\begin{array}{rr} 4 & 1 \\ -2 & 1 \end{array}\right]$$

Use a graphing utility with matrix capabilities or a computer software program to find the eigenvalues of the matrix. $$\left[\begin{array}{rrr} \frac{1}{2} & 0 & 5 \\ -2 & \frac{1}{5} & \frac{1}{4} \\ 0 & 0 & 3 \end{array}\right]$$
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