= Solution
Write $q^{(0)}=\sum_i c_iq_i$ with $c_1\ne0$. After $k$ multiplications, the tangent of the angle to $q_1$ is at most
$$
\frac{\sqrt{\sum_{i\ge2}c_i^2|\lambda_i|^{2k}}}
{|c_1||\lambda_1|^k}
\le\left|\frac{\lambda_2}{\lambda_1}\right|^k\tan\theta_0.
$$
Since $|\sin\theta_k|\le\tan\theta_k$, the first bound follows. Also
$$
\lambda^{(k)}-\lambda_1
=\sum_{i\ge2}(\lambda_i-\lambda_1)|q_i^Tq^{(k)}|^2.
$$
Bound the coefficient by $\max_{i\ge2}|\lambda_i-\lambda_1|$ and the sum by $\sin^2\theta_k$, then use the squared first estimate to obtain the second bound.
Back to article page
