PR 4 # 22 - Page 2

wblipp · 2006-06-26, 04:50

Unless there is something special going on, it's always the largest eigenvalue.

The eigenvectors form a basis set, so any vector x can be represented as

x = a*e₁ + b*e₂

From which we see that

Aⁿx = a*lambda₁ⁿ*e₁ + b*lambda₂ⁿ*e₂

So the larger eigenvalue will always dominate this process unless there is something special forcing the coefficient to zero. The actual iteration adds a constant at every stage, the vector b in

x_n+1 = Ax_n+b

so even if x₀ is aligned with the smaller eigenvector, x₁ will have a component of the larger eigenvector unless b is also a eigenvector for the smaller eigenvalue. (This problem actually picks up the component in x₁ because x₀ is the zero vector).

You can concoct special scenarios with lamda=-1 so that the multiplication and addition cancel each other, but that's about all that can get in the way of the largest eigenvalue dominating the process.

2006-06-26, 04:50	#12
wblipp "William" May 2003 New Haven 2·7·13² Posts	Unless there is something special going on, it's always the largest eigenvalue. The eigenvectors form a basis set, so any vector x can be represented as x = ae₁ + be₂ From which we see that Aⁿx = alambda₁ⁿe₁ + blambda₂ⁿe₂ So the larger eigenvalue will always dominate this process unless there is something special forcing the coefficient to zero. The actual iteration adds a constant at every stage, the vector b in x_n+1 = Ax_n+b so even if x₀ is aligned with the smaller eigenvector, x₁ will have a component of the larger eigenvector unless b is also a eigenvector for the smaller eigenvalue. (This problem actually picks up the component in x₁ because x₀ is the zero vector). You can concoct special scenarios with lamda=-1 so that the multiplication and addition cancel each other, but that's about all that can get in the way of the largest eigenvalue dominating the process.