Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 19E from Chapter 5.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 19E
Chapter:
Problem:
Let A be an n × n matrix, and suppose A has n real eigenvalues repeated according to multiplicities, so that 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.)
Step-by-Step Solution
Given Information
We are given with an $n \times n$ matrix such that \[ \operatorname { det } ( A - \lambda I ) = \left( \lambda _ { 1 } - \lambda \right) \left( \lambda _ { 2 } - \lambda \right) \cdots \left( \lambda _ { n } - \lambda \right) \] We have to explain why determinant of A is the product of the $n$ eigenvalues of A.
Step-1:
Substitute $\lambda = 0$ in above equation \[\begin{array}{l} det (A - (0)I) = \left( {{\lambda _1} - 0} \right)\left( {{\lambda _2} - 0} \right) \ldots \left( {{\lambda _n} - 0} \right)\\ = \left( {{\lambda _1}} \right)\left( {{\lambda _2}} \right) \ldots \left( {{\lambda _n}} \right) \end{array}\] Hence, the characteristic polynomial is:
The determinant of A is the product of n real eigenvalues.
We are given with an $n \times n$ matrix such that \[ \operatorname { det } ( A - \lambda I ) = \left( \lambda _ { 1 } - \lambda \right) \left( \lambda _ { 2 } - \lambda \right) \cdots \left( \lambda _ { n } - \lambda \right) \] We have to explain why determinant of A is the product of the $n$ eigenvalues of A.
Step-1:
Substitute $\lambda = 0$ in above equation \[\begin{array}{l} det (A - (0)I) = \left( {{\lambda _1} - 0} \right)\left( {{\lambda _2} - 0} \right) \ldots \left( {{\lambda _n} - 0} \right)\\ = \left( {{\lambda _1}} \right)\left( {{\lambda _2}} \right) \ldots \left( {{\lambda _n}} \right) \end{array}\] Hence, the characteristic polynomial is:
The determinant of A is the product of n real eigenvalues.