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 5.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 19E
Chapter:
Problem:
find one eigenvalue, with no calculation. Justify your answer.
Step-by-Step Solution
Given Information
We are given that a matrix A\[A = \left[ {\begin{array}{*{20}{l}}1&2&3\\1&2&3\\1&2&3\end{array}} \right]\]We have to find an eigenvalue without any calculation
Step-1:
We can see that all the rows of given matrix are identical, hence the determinant of the matrix A is zero.
We can also write that \[\begin{array}{l}\det A = 0\\\det \left( {A - 0I} \right) = 0\end{array}\]By comparing the standard characteristic equation, having $\lambda$ as eigenvalue, \[\det \left( {A - \lambda I} \right) = 0\]We can see that 0 is an eigenvalue of the matrixTherefore,
One of the eigenvalue is 0
We are given that a matrix A\[A = \left[ {\begin{array}{*{20}{l}}1&2&3\\1&2&3\\1&2&3\end{array}} \right]\]We have to find an eigenvalue without any calculation
Step-1:
We can see that all the rows of given matrix are identical, hence the determinant of the matrix A is zero.
We can also write that \[\begin{array}{l}\det A = 0\\\det \left( {A - 0I} \right) = 0\end{array}\]By comparing the standard characteristic equation, having $\lambda$ as eigenvalue, \[\det \left( {A - \lambda I} \right) = 0\]We can see that 0 is an eigenvalue of the matrixTherefore,
One of the eigenvalue is 0