 Home

Codex Wiki

List of universities

University of Cambridge

Course of the University of Cambridge

Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2022

Paper 3

An orthonormal eigenbasis is
$$
\mathbf v_9=\frac1{\sqrt3}(1,1,1)^T,
\qquad
\mathbf v_6=\frac1{\sqrt6}(1,-2,1)^T,
\qquad
\mathbf v_2=\frac1{\sqrt2}(1,0,-1)^T.
$$
The starting vector has the <orthogonal decomposition by a closed subspace>[orthogonal decomposition]
$$
\mathbf x_0
=0\mathbf v_9+\frac{\sqrt3}{2}\mathbf v_6+\frac12\mathbf v_2.
$$
Hence its component in the dominant eigenspace is exactly zero. Normalizing $A^k\mathbf x_0$ gives
$$
\boxed{
\mathbf x_k
=
\frac{
6^k(1,-2,1)^T+2^k(1,0,-1)^T
}{
\sqrt2\sqrt{3\cdot36^k+4^k}
}
}.
$$
Since the two remaining eigenvectors are orthogonal,
$$
\boxed{
r(\mathbf x_k)
=\frac{18\cdot36^k+2\cdot4^k}
{3\cdot36^k+4^k}
}.
$$
Therefore
$$
\boxed{\lim_{k\to\infty}r(\mathbf x_k)=6}.
$$
The limit is not the leading eigenvalue $9$ because the nonorthogonality assumption in part (a) fails. On the <active spectrum of the power method>, the leading eigenvalue is $6$ and the next is $2$; indeed
$$
r(\mathbf x_k)-6
=-\frac{4\cdot4^k}{3\cdot36^k+4^k}
=O(3^{-2k}),
$$
in agreement with the squared spectral-ratio estimate.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 40C / b / Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 40C / b / Solution

 Ancestors (11)

 Discussion (0)

 Articles by others on the same topic (0)

Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 40C / b / Solution

 Ancestors (11)

 Discussion (0)  Subscribe (1)

 Articles by others on the same topic (0)

 Discussion (0)