Paper 2, Section II, A

Numerical Analysis | Part II, 2011

Let ARn×nA \in \mathbb{R}^{n \times n} be a real matrix with nn linearly independent eigenvectors. The eigenvalues of AA can be calculated from the sequence x(k),k=0,1,x^{(k)}, k=0,1, \ldots, which is generated by the power method

x(k+1)=Ax(k)Ax(k),x^{(k+1)}=\frac{A x^{(k)}}{\left\|A x^{(k)}\right\|},

where x(0)x^{(0)} is a real nonzero vector.

(i) Describe the asymptotic properties of the sequence x(k)x^{(k)} in the case that the eigenvalues λi\lambda_{i} of AA satisfy λi<λn,i=1,,n1\left|\lambda_{i}\right|<\left|\lambda_{n}\right|, i=1, \ldots, n-1, and the eigenvectors are of unit length.

(ii) Present the implementation details for the power method for the setting in (i) and define the Rayleigh quotient.

(iii) Let AA be the 3×33 \times 3 matrix

A=λI+P,P=(000100010)A=\lambda I+P, \quad P=\left(\begin{array}{lll} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{array}\right)

where λ\lambda is real and nonzero. Find an explicit expression for Ak,k=1,2,3,A^{k}, k=1,2,3, \ldots

Let the sequence x(k)x^{(k)} be generated by the power method as above. Deduce from your expression for AkA^{k} that the first and second components of x(k+1)x^{(k+1)} tend to zero as kk \rightarrow \infty. Further show that this implies Ax(k+1)λx(k+1)0A x^{(k+1)}-\lambda x^{(k+1)} \rightarrow 0 as kk \rightarrow \infty.

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