Could someone give me a hint on how I should show that the power method applied to $(A - \lambda_1 I)q$ leads to an approximation of $\lambda_2$, where $q$ is an arbitrary vector, assuming that we know $\lambda_1$ and $v_1$ (the eigenvector corresponding to $\lambda_1$).
I think I can write $q = a_1v_1 + a_2 v_2 + \ldots, a_nv_n$ and use this somehow, but unfortunately I haven't been able to get any useful results..
Edit: Apparently, applying the power method to $(A - \lambda_1I)q$ does in fact lead to approximating $\lambda_2$; at least this article says it does on page $3$.
Furthermore, I think I can rewrite $(A - \lambda_1I)q$ as $$(A - \lambda_1I)q = a_2\lambda_2v_2 + a_3\lambda_3v_3 + \ldots + a_n\lambda_nv_n - \lambda_1(a_2v_2 + a_3v_3 + \ldots + a_nv_n) =\\-(a_2v_2 + \ldots + a_nv_n) + (\dfrac{\lambda_2}{\lambda_1}a_2v_2 + \dfrac{\lambda_3}{\lambda_1}a_3v_3 + \ldots)$$ Is this the right way to go?
Your reference is suggesting that you construct a vector which lies in $\mathrm{span}(\{ v_2,\dots,v_n \}) \setminus \mathrm{span}(\{ v_3,\dots,v_n \})$. That is, it should be of the form $\sum_{i=2}^n c_i v_i$ where $c_2 \neq 0$. In principle this condition can be checked by considering the left eigenvectors, like your reference suggests. But this is not usually done except when the eigenvectors of $A$ are orthogonal.
Once you have constructed the initial vector, you still run power iteration with $A$, not with $A-\lambda_1 I$. You construct the initial vector by using $A-\lambda_1 I$ to "kill" the component in the direction of $v_1$ of whatever $x_0$ you started with. Then your sequence of iterates is $A^n(A-\lambda_1 I)x_0,n=0,1,\dots$. This will indeed converge to an eigenvector of $A$ with eigenvalue $\lambda_2$ (with the caveats given in the reference).
Power iteration with $A-\lambda_1 I$ will generally work, providing you with a different eigenvector of $A$ than the one you already found. But the corresponding eigenvalue will almost never be $\lambda_2$. In fact it will be $\min_i \lambda_i$ if $\lambda_1>0$ or $\max_i \lambda_i$ if $\lambda_1<0$.