Explanation The provided matrix is A = ( 2 1 1 2 ) . Consider the matrix A , A = ( 2 1 1 2 ) Since, the characteristic equation for the matrix A is | ( A − λ I ) | = 0 where I is the identity matrix. Substitute the values of matrix A and I in above expression. det ( A − λ Ι ) = 2 − λ 1 1 2 − λ = λ 2 − 4 λ + 3 = ( λ − 3 ) ( λ − 1 ) Now, equating above expression to 0 gives ( λ − 3 ) ( λ − 1 ) = 0 , So, the Eigen values are λ 1 = 1 and λ 2 = 3 . Now, to obtain the Eigen vector K 1 corresponding to the Eigen value λ 1 = 1 , solve the equation ( A − λ 1 I ) K 1 = 0 where, K 1 = ( x 1 y 1 z 1 ) . Substituting the values of A , λ 1 and K 1 in the above equation gives, [ ( 2 1 1 2 ) − ( 1 0 0 1 ) ] ⋅ ( x 1 y 1 ) = ( 0 0 ) ( 2 − 1 1 1 2 − 1 ) ( x y ) = ( 0 0 ) ( 1 1 1 1 ) ( x y ) = ( 0 0 ) ( x + y x + y ) = ( 0 0 ) Which on further solving gives, x + y = 0 x + y = 0 Which means y = − x Therefore, the Eigen vector corresponding to λ 1 = 1 is, K 1 = ( − 1 1 ) Similarly, to obtain the Eigen vector corresponding to λ 2 = 3 [ ( 2 1 1 2 ) − ( 3 0 0 3 ) ] ⋅ ( x 1 y 1 ) = ( 0 0 ) ( 2 − 3 1 1 2 − 3 ) ( x y ) = ( 0 0 ) ( − 1 1 1 − 1 ) ( x y ) = ( 0 0 ) ( − x + y x − y ) = ( 0 0 ) Now, equate the above matrix to zero − x + y = 0 x − y = 0 Which on solving gives x = y Therefore, the Eigen vector corresponding to λ 2 = 3 is K 2 = ( 1 1 ) Hence, the Corresponding eigenvectors are, K 1 = ( − 1 1 ) and K 2 = ( 1 1 ) Let P denote a matrix whose column are eigenvectors K 1 , K 2 , … K n corresponding to distinct Eigen values λ 1 , λ 2 , … λ n of an n × n matrix A

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Author: Dennis G. Zill

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Chapter 10.5, Problem 26E

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