The Chemistry Maths Book, Second Edition

(Grace) #1

556 Chapter 19The matrix eigenvalue problem


20.The Hückel Hamiltonian matrix of butadiene is


Find (i)the eigenvalues (in terms of αand β), (ii)the orthonormal eigenvectors.


You may find the following relations useful: (the ‘golden section’,


see Section 7.2), ,,φ


2

1 − 111 = 1 φ,φ


2

1 − 1 φ 1 = 1 1.


21.The Hückel Hamiltonian matrix of cyclopropene is


(i) Show that the eigenvalues areE


1

1 = 1 α 1 + 12 β,E


2

1 = 1 E


3

1 = 1 α 1 − 1 β. (ii) Find the normalized


eigenvector belonging to eigenvalueE


1

. (iii) Show that an eigenvector belonging to the


doubly-degenerate eigenvalueα 1 − 1 βhas componentsx


1

, x


2

, x


3

that satisfyx


1

1 + 1 x


2

1 + 1 x


3

1 = 10.


(iv) Find two orthonormal eigenvectors corresponding to eigenvalueα 1 − 1 β(you may


find the results of Examples 19.4(ii) and 19.8 useful).



  1. (i)Show that a square matrix Aand its transposeA


T

have the same set of eigenvalues.


(ii) Show that the following two equations are equivalent:


A


T

y 1 = 1 λy, y


T

A 1 = 1 λy


T

The eigenvectors ofA


T

are in general different from those of A(unless Ais symmetric).


The vector y is sometimes called a left-eigenvectorof A, and an ‘ordinary’ eigenvector x


of Ais then called a right-eigenvector.


(iii)Find the eigenvalues and corresponding normalized right- and left-eigenvectors of


.


Section 19.4


23.For the matrix


of Exercise 8, construct (i) the matrix Xof the eigenvectors and (ii) the diagonal matrixD


of the eigenvalues.(iii) Show thatAX 1 = 1 DX.


24.Repeat Exercise 23 for the matrix of Exercise 9.


25.For the matrix of Exercise 7,


(i)construct the matrix Xof the eigenfunctions ofA, and find its inverse,X


− 1

,


(ii)calculateD 1 = 1 X


− 1

AXand confirm that Dis the diagonal matrix of the eigenvalues


ofA.


A=










31


13


030


303


030














A=














120


210


021


A=










32


02


αββ


βα β


ββα














φφ−=15121()− =
φ

2

=+() 532


φ=+()512


αβ


βα β


βα β


βα


00


0


0


00















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