Paper 4, Section II, B

Numerical Analysis | Part II, 2016

(a) Describe an implementation of the power method for determining the eigenvalue of largest modulus and its associated eigenvector for a matrix that has a unique eigenvalue of largest modulus.

Now let AA be a real n×nn \times n matrix with distinct eigenvalues satisfying λn=λn1\left|\lambda_{n}\right|=\left|\lambda_{n-1}\right| and λn>λi,i=1,,n2\left|\lambda_{n}\right|>\left|\lambda_{i}\right|, i=1, \ldots, n-2. The power method is applied to AA, with an initial condition x(0)=i=1nciwi\mathbf{x}^{(0)}=\sum_{i=1}^{n} c_{i} \mathbf{w}_{i} such that cn1cn0c_{n-1} c_{n} \neq 0, where wi\mathbf{w}_{i} is the eigenvector associated with λi\lambda_{i}. Show that the power method does not converge. Explain why x(k),x(k+1)\mathbf{x}^{(k)}, \mathbf{x}^{(k+1)} and x(k+2)\mathbf{x}^{(k+2)} become linearly dependent as kk \rightarrow \infty.

(b) Consider the following variant of the power method, called the two-stage power method, applied to the matrix AA of part (a):

  1. Pick x(0)Rn\mathbf{x}^{(0)} \in \mathbb{R}^{n} satisfying x(0)=1\left\|\mathbf{x}^{(0)}\right\|=1. Let 0<ε10<\varepsilon \ll 1. Set k=0k=0 and x(1)=Ax(0)\mathbf{x}^{(1)}=A \mathbf{x}^{(0)}.

  2. Calculate x(k+2)=Ax(k+1)\mathbf{x}^{(k+2)}=A \mathbf{x}^{(k+1)} and calculate α,β\alpha, \beta that minimise

f(α,β)=x(k+2)+αx(k+1)+βx(k).f(\alpha, \beta)=\left\|\mathbf{x}^{(k+2)}+\alpha \mathbf{x}^{(k+1)}+\beta \mathbf{x}^{(k)}\right\| .

  1. If f(α,β)εf(\alpha, \beta) \leqslant \varepsilon, solve λ2+αλ+β=0\lambda^{2}+\alpha \lambda+\beta=0 and let the roots be λ1\lambda_{1} and λ2\lambda_{2}. They are accepted as eigenvalues of AA, and the corresponding eigenvectors are estimated as x(k+1)λ2x(k)\mathbf{x}^{(k+1)}-\lambda_{2} \mathbf{x}^{(k)} and x(k+1)λ1x(k)\mathbf{x}^{(k+1)}-\lambda_{1} \mathbf{x}^{(k)}.

  2. Otherwise, divide x(k+2)\mathbf{x}^{(k+2)} and x(k+1)\mathbf{x}^{(k+1)} by the current value of x(k+1)\left\|\mathbf{x}^{(k+1)}\right\|, increase kk by 1 and return to Step 1\mathbf{1}.

Explain the justification behind Step 2 of the algorithm.

(c) Let n=3n=3, and suppose that, for a large value of kk, the two-stage power method yields the vectors

yk=x(k)=(111),yk+1=Ax(k)=(234),yk+2=A2x(k)=(246)\mathbf{y}_{k}=\mathbf{x}^{(k)}=\left(\begin{array}{l} 1 \\ 1 \\ 1 \end{array}\right), \quad \mathbf{y}_{k+1}=A \mathbf{x}^{(k)}=\left(\begin{array}{c} 2 \\ 3 \\ 4 \end{array}\right), \quad \mathbf{y}_{k+2}=A^{2} \mathbf{x}^{(k)}=\left(\begin{array}{c} 2 \\ 4 \\ 6 \end{array}\right)

Find two eigenvalues of AA and the corresponding eigenvectors.

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