Do complex linear automorphisms always allow a basis of orthogonal eigenvectors? I want to understand why the following lemma, given without proof in an article (and a reference which does not seem to answer it at all) holds:
Lemma: If $f \in \text{GL}_n(\mathbb{C})$ has an algebraically simple real eigenvalue $\rho$ such that for all other eigenvalues $\lambda \in \text{spec}(f)$ we have $|\lambda| < \rho$ and if $y$ is an eigenvector of $\rho$ and if $\mathbb{C}^n = \mathbb{C}y \oplus W$ with $f(W) = W$, then for $x = \alpha y + w$ with $\alpha \in \mathbb{C}$ and $w \in W$ we have
$$\lim_{n \to \infty} \frac{1}{\rho^n} f^n x = \alpha y.$$
My attempts: We have
$$\frac{1}{\rho^n}f^nx = \frac{\alpha}{\rho^n}f^n y + \frac{1}{\rho^n}f^nw = \alpha y + \frac{1}{\rho^n}f^n w.$$
This converges to $\alpha y$ iff $\lim_{n \to \infty}\frac{1}{\rho^n}f^n w = 0$. This is for example true in the case that $f$ allows a basis of orthogonal eigenvectors, since then $\|f^nw\| \leq |\rho'|^n \cdot \|w\|$, where $\rho'$ is an eigenvalue such that $|\rho'|$ has the second biggest size among all eigenvectors. Then as $\left( \frac{\rho'}{\rho} \right)^{n} \to 0$ we are done. This leads to the following question:
Question: Do complex linear automorphisms always allow a basis of orthogonal eigenvectors?
I also have a second question (sorry, linear algebra was a long time ago...):
Question: Does a subspace $W$ as in the Lemma always exist if $f$ has an algebraically simple eigenvalue $\rho$ such that all other eigenvalues have smaller absolute value?
It would also be nice if you knew references.
Best regards,
Leon
 A: No, complex operators do not necessarily have an orthonormal basis of eigenvectors nor even a non-orthnormal basis of eigenvectors. In fact, an operator will have an orthonormal basis of eigenvectors if and only if it is normal (this is the context of the complex spectral theorem).
However, a useful replacement for orthogonal diagonalizability which also solves your problem is given by the Jordan form of an operator. Let $f \in \operatorname{GL}_n(\mathbb{C})$ be an operator with an algebraically simple, real, positive eigenvalue $\rho > 0$ and denote the other eigenvalues of $f$ by $\lambda_1, \dots, \lambda_k$. You are given that $|\lambda_i| < \rho$ for all $1 \leq i \leq k$. Denote by $V_{\rho}, V_{\lambda_1}, \dots, V_{\lambda_k}$ the generalized eigenspaces of $f$. We always have a direct sum decomposition
$$ \mathbb{C}^n = V_{\rho} \oplus V_{\lambda_1} \oplus \dots \oplus V_{\lambda_k} $$
and each generalized eigenspace is $f$-invariant. In addition, since $\rho$ is algebraically simple, we have $\dim V_{\rho} = 1$ (and this is a true eigenspace). Hence, if you set $W = V_{\lambda_1} \oplus \dots V_{\lambda_k}$ and choose a non-zero vector $y \in V_{\rho}$ you get a direct sum decomposition $\mathbb{C}^n = \mathbb{C}y \oplus W$ with $f(W) \subseteq W$. If $f$ is invertible (equivalently, if $|\lambda_i| > 0$ for all $1 \leq i \leq k$) then $f(W) = W$.
Next, given $x \in \mathbb{C}^n$, decompose it as $x = \alpha y + w$ where $w \in W$ and then
$$ \frac{f^k(x)}{\rho^k} = \alpha y + \frac{f^k(w)}{\rho^k}. $$
Like you noted, to show that this goes to $\alpha y$ as $k \to \infty$, it is enough to show that $\frac{f^k(w)}{\rho^k} \to 0$. The Jordan normal form tells us that we can write $f|_{W} = S + N$ where $S \colon W \rightarrow W$ is diagonalizable with eigenvalues $\lambda_1, \dots, \lambda_k$ and $N$ is nilpotent such that $SN = NS$. Since $\dim W = n - 1$, we have $N^{n-1} = 0$. Hence by the binomial theorem (using $NS = SN$) we have
$$ f^k(w) = (S + N)^k(w) = \left( \sum_{i=0}^{n - 1} {k \choose i} S^{k - i} N^i \right)(w) = \\
 S^k(w) + {k \choose 1} S^{k-1}(Nw) + \dots + {k \choose n - 1} S^{k - n + 1}(N^{n-1}(w)).$$
Finally, show that ${k \choose r} \frac{S^k(w')}{\rho^k} \to 0$ for all $r$ and $w' \in W$ using the fact that $S$ is diagonalizable with eigenvalues of modulus less than $\rho$ by choosing a basis of eigenvectors of $S$ of $W$ and apply this to each of the $n$ summands above.
