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Chapter 5: Q5.2-19E (page 267)

Question 19: Let \(A\) be an \(n \times n\) matrix, and suppose A has \(n\) real eigenvalues, \({\lambda _1},...,{\lambda _n}\), repeated according to multiplicities, so that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) . Explain why \(\det A\) is the product of the n eigenvalues of A. (This result is true for any square matrix when complex eigenvalues are considered.)

Short Answer

Expert verified

The \(\det A\) is the product of \(n\) eigenvalues of A.

Step by step solution

01

Definition of a multiplicity of an eigenvalue

In particular, the multiplicity of an eigenvalue \(\lambda \) represents its multiplication as a root of the characteristic equation.

02

Explain why \(\det A\) is the product of \(n\) eigenvalues of A

Consider \(A\) as an \(n \times n\) matrix and let \(A\) has \(n\) real eigenvalues \({\lambda _1}, \ldots ,{\lambda _n}\), repeated from the multiplicities.

For all \(\lambda \), it is true that \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\). Let \(\lambda = 0\), so \(\det A = {\lambda _1},{\lambda _2}, \ldots ,{\lambda _n}\).

It is deduced that \(\det A = {\lambda _1},{\lambda _2}, \ldots ,{\lambda _n}\) because the equation \(\det \left( {A - \lambda I} \right) = \left( {{\lambda _1} - \lambda } \right)\left( {{\lambda _2} - \lambda } \right) \ldots \left( {{\lambda _n} - \lambda } \right)\) is true for all \(\lambda \).

Thus, \(\det A\) is the product of \(n\) eigenvalues of A.

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Most popular questions from this chapter

Let \(A\) be a real \(2 \times 2\) matrix with a complex eigenvalue \(\lambda = a - bi\)\(\left( {b \ne 0} \right)\) and an associated eigenvector \({\bf{v}}\) in \({\mathbb{}^2}\).

  1. Show that \(A({\mathop{\rm Re}\nolimits} {\bf{v}}) = a{\mathop{\rm Re}\nolimits} {\bf{v}} + b{\mathop{\rm Im}\nolimits} {\bf{v}}\) and \(A({\mathop{\rm Im}\nolimits} {\bf{v}}) = - b{\mathop{\rm Re}\nolimits} {\bf{v}} + a{\mathop{\rm Im}\nolimits} {\bf{v}}\). (Hint: Write \({\bf{v}} = {\mathop{\rm Re}\nolimits} {\bf{v}} + i{\mathop{\rm Im}\nolimits} {\bf{v}}\), and compute \(A{\bf{v}}\).)
  2. Verify that if \(P\) and \(C\) are given as in Theorem 9, then \(AP = PC\)

Question: Repeat Exercise 35, assuming u and v are eigenvectors of A that correspond to eigenvalues -1 and 3, respectively.

In Exercises 7鈥�12, use Example 6 to list the eigenvalues of\(A\). In each case, the transformation\({\rm{x}} \mapsto A{\rm{x}}\)is the composition of a rotation and a scaling. Give the angle\(\varphi \)of the rotation, where\( - \pi < \varphi \le \pi \)and give the scale\(r\).

11.\(\left( {\begin{aligned}{}{\,\,\,.1}&{}&{.1}\\{ - .1}&{}&{.1}\end{aligned}} \right)\)

Question: A is a \({\bf{7}} \times {\bf{7}}\) matrix with three eigenvalues. One eigenspace is two-dimensional and one of the other eigenspaces is three-dimensional. Is it possible that A is not diagonalizable? Justify your answer.

In Exercises 9鈥�16, find a basis for the eigenspace corresponding to each listed eigenvalue.

10. \(A = \left( {\begin{array}{*{20}{c}}{10}&{ - 9}\\4&{ - 2}\end{array}} \right)\), \(\lambda = 4\)
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