Show that this iterative Richardson iteration may diverge I need a check on the following problem:

Let $A$ a nonsingular matrix with real eigenvalues, and consider the iterative scheme $$x_{k+1} = x_k + \alpha (b- Ax_k)$$ for $\alpha \ne 0$.
i) Assume that $A$ has both negative and real eigenvalues. Show that for every $\alpha \ne 0$ there exists $x_0$ s.t. $\{ x_k\}_k$ does not converge
ii) Assume that $A$ has only positive eigenvalues. Find conditions on $\alpha$ s.t. the method converges for every $x_0$. Find also the value of $\alpha$ that minimize the spectral radius.


I have big problems with the first point.
i) I notice that the iteration matrix is $R=I-\alpha A$. Therefore, the eigenvalues are $\lambda (R)=1-\alpha\lambda$. The requirement to have convergence is that $\sigma(R)<1$, and so it must be $$|1-\alpha \lambda|<1$$ which implies,as $\lambda \in \mathbb{R}$: $$\frac{2}{\alpha \lambda_i}>1$$ (it is well defined, as $\det(A)= \prod \lambda_i  \ne 0$ and so each $\lambda_i \ne 0$)
The fact is that we don't know anything more about that quotient. So, if the sign of the eigevalues is not constant (as it could be from the assumptions), the method will diverge.
ii) Here I just imposed that for every $i$: $$|1-\lambda_i \alpha|<1$$ i.e. $$\alpha \in (0,\frac2\lambda_i)$$ Assume that $\lambda_1 \geq \lambda_2 \geq \lambda_n \geq 0$
so the last condition becomes $$\alpha \in (0,\frac2\lambda_1)$$
Then, in order to minimize the spectral radius, I impose $$1-\alpha \lambda_n = -(1-\alpha \lambda_1)$$ therefore it follows $$\alpha=\frac{2}{\lambda_1 + \lambda_n}$$ minimizes the spectral radius
Is everything okay?
 A: I think it may be useful to take a step back to see exactly where the spectral radious criteria comes from.
Suppose $x$ is the exact solution satisfying $Ax = b$, if we define the error on the $k$-th iteration as $e_k = x_k-x$, remember that
$$e_{k+1} = (I -\alpha A)e_k = Re_k$$
So by setting $e_0 = x_0-x$, the error in a given iteration $k \in \mathbb{N}$ simplifies to $e_k = R^k e_0$.
It can be shown that  $R^k \rightarrow 0$ as $k\rightarrow\infty$ if and only if all eigenvalues of $R$ have absolute value strictly less than $1$, so the spectral radius criteria is necessary and sufficient to have convergence for any given $e_0$.
Perhaps the confusion is here: even if $R^k \nrightarrow 0$, the method still converges for some choices of $x_0$. As an example, for any $R$, $e_0 \in ker(R) \implies e_1 = Re_0 = 0 \implies e_k \rightarrow 0$ as $k\rightarrow \infty$. So to find an $x_0$ that makes the method diverge, the initial choice of $x_0$ has to be more specific.
To get an explicit initial condition that makes the iteration diverge, start by taking an eigenpair $(\lambda_*, v_*)$ from $A$ and notice that since
$$Rv_* = (I-\alpha A)v_* = v_*-\alpha(Av_*) = (1-\alpha \lambda_*)v_*$$
$v_*$ will also be an eigenvector of $R$ with eigenvalue $(1-\alpha\lambda_*)$ associated.
But, as you already found out, $A$ having eigenvalues with different signs implies that $|1- \alpha \lambda_*| \geq 1$ for some $(\lambda_*, v_*)$. So by making $e_0 = v_*$ with $x_0 = v_*+x$,
$$\lim_{k\rightarrow \infty} e_k = \lim_{k \rightarrow \infty} R^k v_* = \lim_{k \rightarrow \infty} {\overbrace{(1-\alpha\lambda)}^{\geq 1}} {}^k v_* \neq 0$$
and therefore divergence is guaranteed.
Your solution to Part ii) looks good to me!
