Linear Algebra

Topic: Method of Powers 395Topic: Method of PowersIn practice, calculating eigenvalues and eigenvectors is a difficult problem. Finding,and solving, the characteristic polynomial of the large matrices often encounteredin applications is too slow and too hard. Other techniques, indirectones that avoid the characteristic polynomial, are used. Here we shall see sucha method that is suitable for large matrices that are ‘sparse’ (the great majorityof the entries are zero).Suppose that the n×n matrix T has the n distinct eigenvalues λ 1 , λ 2 , . . . , λ n .Then R n has a basis that is composed of the associated eigenvectors 〈 ⃗ ζ 1 , . . . , ⃗ ζ n 〉.For any ⃗v ∈ R n , where ⃗v = c 1⃗ ζ1 + · · · + c n⃗ ζn , iterating T on ⃗v gives these.T⃗v = c 1 λ 1⃗ ζ1 + c 2 λ 2⃗ ζ2 + · · · + c n λ n⃗ ζnT 2 ⃗v = c 1 λ 2 1 ⃗ ζ 1 + c 2 λ 2 2 ⃗ ζ 2 + · · · + c n λ 2 n ⃗ ζ nT 3 ⃗v = c 1 λ 3 1 ⃗ ζ 1 + c 2 λ 3 2 ⃗ ζ 2 + · · · + c n λ 3 n ⃗ ζ n..T k ⃗v = c 1 λ k 1 ⃗ ζ 1 + c 2 λ k 2 ⃗ ζ 2 + · · · + c n λ k n ⃗ ζ nIf one of the eigenvaluse, say, λ 1 , has a larger absolute value than any of theother eigenvalues then its term will dominate the above expression. Put anotherway, dividing through by λ k 1 gives this,T k ⃗vλ k 1= c 1⃗ ζ1 + c 2λ k 2λ k 1λ⃗ζ k n2 + · · · + c n⃗ζλ k n1and, because λ 1 is assumed to have the largest absolute value, as k gets largerthe fractions go to zero. Thus, the entire expression goes to c 1ζ1 ⃗ .That is (as long as c 1 is not zero), as k increases, the vectors T k ⃗v willtend toward the direction of the eigenvectors associated with the dominanteigenvalue, and, consequently, the ratios of the lengths ‖ T k ⃗v ‖/‖ T k−1 ⃗v ‖ willtend toward that dominant eigenvalue.For example (sample computer code for this follows the exercises), becausethe matrix( ) 3 0T =8 −1is triangular, its eigenvalues are just the entries on the diagonal, 3 and −1.Arbitrarily taking ⃗v to have the components 1 and 1 gives(⃗v) (T⃗v) (T 2 ⃗v)· · ·(T 9 ⃗v) (T 10 ⃗v)1 3 919 683 59 049· · ·1 7 1739 367 118 097and the ratio between the lengths of the last two is 2.999 9.Two implementation issues must be addressed. The first issue is that, insteadof finding the powers of T and applying them to ⃗v, we will compute ⃗v 1 as T⃗v and