If a diagonalisable matrix has a simple eigenvalue of strictly largest modulus and the start vector has a nonzero component in that eigendirection, power iteration converges projectively at the eigenvalue-ratio rate

By [L1], the chosen eigenvectors form a basis, so for every $k\ge 0$, $A^k{x}_0={\lambda }_1^k(c_1{v}_1+{\sum }_{i=2}^n{c}_i{\left(\frac{{\lambda }_i}{{\lambda }_1}\right)}^k{v}_i)\mathrm{.}$

The strict modulus inequality and $n\ge 2$ imply ${\lambda }_1\mathrm{\ne }0$. In the eigenbasis, the $v_1$-coefficient of $A^k{x}_0$ is $c_1{\lambda }_1^k\mathrm{\ne }0$, so $A^k{x}_0\mathrm{\ne }0$ for every $k\ge 0$. Therefore the iteration of [L2] is defined at every step and has the same direction as $A^k{x}_0$. Moreover, because $\mathrm{\mid }{\lambda }_i\mathrm{/}{\lambda }_1\mathrm{\mid }<1$ for $i\ge 2$, the bracket in step 1.1 tends to $c_1{v}_1$.

Let ${\alpha }_k$ cancel the phase of ${\lambda }_1^k{c}_1$. Then step 1.1 gives ${\alpha }_k{x}_k=\frac{v_1+{\sum }_{i=2}^n(c_i\mathrm{/}{c}_1)({\lambda }_i\mathrm{/}{\lambda }_1{)}^k{v}_i}{{\parallel v_1+{\sum }_{i=2}^n(c_i\mathrm{/}{c}_1)({\lambda }_i\mathrm{/}{\lambda }_1{)}^k{v}_i\parallel }_2}\mathrm{.}$ The numerator differs from $v_1$ by $O(\mathrm{\mid }{\lambda }_2\mathrm{/}{\lambda }_1{\mathrm{\mid }}^k)$, so the same is true after normalisation.

Therefore the normalised iterates converge projectively to the eigendirection of $v_1$, and the convergence rate is $O(\mathrm{\mid }{\lambda }_2\mathrm{/}{\lambda }_1{\mathrm{\mid }}^k)$.