Conditions to to have $\rho(I-aM)<1$ Let $M$ be a non-symmetric matrix with complex eigenvalues $m_1,\dots, m_n$. Why, in order to prove $\rho(I-aM)<1$, does it suffice to show that $\mathrm{Re}\{m_j\} > 0$?
Notations: $\rho$ is the spectral radius operator while $\mathrm{Re}\{\cdot\}$ operator returns the real part of a complex number.
 A: To quote the text: "the proof consists of showing that, for $\alpha$ sufficiently small, the hypothesis of Ostrowski's theorem is satisfied". In this case, we want to show that the matrix $B$ is such that, for sufficiently small $\alpha$, $\rho(I - \alpha B) < 1$. To show that this is the case, it suffices to show that the real parts of the eigenvalues of $B$ have strictly positive real part. The proof proceeds to do this, allowing us to reach the desired conclusion.

Claim: If the eigenvalues of $B$ have strictly positive real parts, then there exists some $\bar \alpha > 0$ such that $\rho(I - \alpha B) < 1$ for all $\alpha \in (0,\bar \alpha]$.

Proof: Let $m_k = p_k + iq_k$ denote the $k$th eigenvalue of $B$. The eigenvalues of $I - \alpha B$ will be of the form $\lambda_k = 1 - \alpha m_k$ for $k = 1,\dots,n$. We note that
$$
|1 - \alpha m_k|^2 = (1 - \alpha p_k)^2 + (\alpha q_k)^2 = (p_k^2 + q_k^2)\alpha^2 - 2p_k \alpha + 1.
$$
For each $k$, we may consider the quadratic function $f_k:\Bbb R \to \Bbb R$ defined by
$$
f_k(\alpha) = (p_k^2 + q_k^2)\alpha^2 - 2p_k \alpha + 1.
$$
Each $f_k$ satisfies $f_k(0) = 1$ and achieves a minimum at
$$
\alpha = \bar \alpha_k := \frac{p_k}{p_k^2 + q_k^2} > 0.
$$
Each function $f_k$ is strictly decreasing over the interval $[0,\bar \alpha_k]$. Now, define
$$
\bar \alpha = \min_{k = 1,\dots,n} \bar \alpha_k.
$$
From our observations so far, we can conclude that for each $k = 1,\dots,n$, $f_k(\alpha)$ will be less than $f(0) = 1$ for each $\alpha \in (0,\bar \alpha]$. That is, for each $\alpha \in (0,\bar \alpha]$, the eigenvalues $\lambda_k$ of $I - \alpha B$ will all satisfy $|\lambda_k|^2 < 1$, which is to say that we have $\rho(I - \alpha B) < 1$.
$\hspace{300 pt}\square$
