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**Unformatted text preview: **ubic in the sense that if λJ is an
eigenvalue of A and v(0) is suﬃciently close to the eigenvector qJ , then
v(k +1) − (±qJ ) = O ( v(k ) − (±qJ ) 3 )
and
|λ(k +1) − λJ | = O (|λ(k ) − λJ |3 )
as k → ∞. The ± sign means that at each step k , one or the other choice
of sign is to be taken, and then the indicated bound holds. 7 / 21 Example
Consider the symmetric matrix 211
A = 1 3 1
114
√
and let v(0) = (1, 1, 1) / 3 the initial eigenvector estimate
When Rayleigh quotient iteration is applied to A, the following values
λ(k ) are computed by the ﬁrst 3 iterations:
λ(0) = 5, λ(1) = 5.2131 . . . , λ(2) = 5.214319743184 . . . The actual value of the eigenvalue corresponding to the eigenvector
closest to v(0) is λ = 5.214319743377
After three iterations, Rayleigh quotient iteration has produced a
result accurate to 10 digits
With three more iterations, the solution increases this ﬁgure to about
270 digits, if our machine precision were high enough
8 / 21 Operation counts
Flops (ﬂoating point operations): addition, subtraction,
multiplication, division, or square root counts as one ﬂop
Suppose A ∈ IRm×m is a full matrix, each step of power iteration
involves a matrix-vector multiplication, requiring O (m2 ) ﬂops
Each step of inverse iteration involves the solution of a linear system,
which might seem to require O (m3 ) ﬂops, but this can be reduced to
O (m2 ) if the matrix is processed in advance by LU or QR factorization
In Rayleigh quotient iteration, the matrix to be inverted changes at
each step, and beating O (m3 ) ﬂops per step
These ﬁgures improve greatly if A is tridiagonal, and all three
iterations requires just O (m) ﬂops per step
For non-symmetric matrices, we must deal with Hessenberg instead of
tridiagona...

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