Why do we have $\|q_{k-1}\|_2^2 = 1$?

Question

I am reading the following text about the power method:

The Power Method: $q_0\in\mathbb{C}^n$ is given

for $k = 1,2,\ldots$ do $$z_k = Aq_{k-1}\\q_k = z_k/\|z_k\|_2\\\lambda^{(k)} = \overline{q}_{k-1}^Tz_k$$ end for

It is easy to see that if $q_{k-1}$ is an eigenvector corresponding to $\lambda_j$ then $$\lambda^{(k)} = q_{k-1}^TAq_{k-1} = \lambda_j\|q_{k-1}\|_2^2 = \lambda_j$$

Question: Why is this so easy to see? I'm probably missing something extremely obvious, but I don't see why $\|q_{k-1}\|_2^2 = 1$. Eigenvectors don't have to be unit vectors right?

Edit: Furthermore, what's the difference between $\overline{q}_{k-1}$ and $q_{k-1}$? Is this just a typo? Or does it refer to the normalized vector?

Thanks!

Answer 1

There is a normalization step of $$q_k = \frac{z_k}{\|z_k\|_2}$$

$\bar{q}_{k-1}$ is the conjugate transpose of $q_{k-1}$.