Endomorphism/K/Powers/Zero convergence/Fact/Proof

(1) implies (2). Let ${\displaystyle {}v\in V}$. We can work with an arbitrary norm on ${\displaystyle {}V}$ and on the endomorphism space, for example, with the maximum norm. Because of

${\displaystyle {}\Vert {\varphi ^{n}(v)}\Vert \leq \Vert {\varphi ^{n}}\Vert _{\rm {max}}\cdot \Vert {v}\Vert \,,}$

the sequence ${\displaystyle {}\varphi ^{n}(v)}$ converges to ${\displaystyle {}0}$. From (2) to (3) is clear. If (3) holds, and

${\displaystyle {}v=a_{1}v_{1}+\cdots +a_{m}v_{m}\,}$

is a linear combination, then

${\displaystyle {}\varphi ^{n}{\left(a_{1}v_{1}+\cdots +a_{m}v_{m}\right)}=a_{1}\varphi ^{n}(v_{1})+\cdots +a_{m}\varphi ^{n}(v_{m})\,,}$

and the convergence of the sequences ${\displaystyle {}\varphi ^{n}(v_{j})}$ to ${\displaystyle {}0}$ implies the convergence of this sum sequence to ${\displaystyle {}0}$. From (2) or (3) to (4). We may assume ${\displaystyle {}{\mathbb {K} }=\mathbb {C} }$: In the real case, we can stick to ${\displaystyle {}V=\mathbb {R} ^{d}}$, and the mapping is given by a real matrix, which we can consider as a complex matrix. The given real generating system is also a complex generating system for ${\displaystyle {}\mathbb {C} ^{d}}$. Let ${\displaystyle {}\lambda }$ be an eigenvalue and ${\displaystyle {}v\in V}$ be an eigenvector of ${\displaystyle {}\lambda }$. Due to the condition,

${\displaystyle {}\varphi ^{n}(v)=\lambda ^{n}v\,}$

converges to ${\displaystyle {}0}$; therefore, ${\displaystyle {}\lambda ^{n}}$ converges to ${\displaystyle {}0}$, and so

${\displaystyle {}\vert {\lambda }\vert <1\,.}$

To get the implication from (4) to (1), we use fact. We have

${\displaystyle {}\varphi ^{n}=\varphi _{\rm {diag}}^{n}+n\varphi _{\rm {diag}}^{n-1}\circ \varphi _{\rm {nil}}+{\binom {n}{2}}\varphi _{\rm {diag}}^{n-2}\circ \varphi _{\rm {nil}}^{2}+{\binom {n}{3}}\varphi _{\rm {diag}}^{n-3}\circ \varphi _{\rm {nil}}^{3}+\cdots +{\binom {n}{k-1}}\varphi _{\rm {diag}}^{n-k+1}\circ \varphi _{\rm {nil}}^{k-1}\,,}$

where ${\displaystyle {}k}$ is the order of nilpotency of the nilpotent part ${\displaystyle {}\varphi _{\rm {nil}}}$. The eigenvalues of ${\displaystyle {}\varphi }$ are according to exercise the eigenvalues of the diagonalizable part ${\displaystyle {}\varphi _{\rm {diag}}}$; we denote them by ${\displaystyle {}z_{j}}$. The summands are of the form

${\displaystyle p(n)\varphi _{\rm {diag}}^{n-s}\circ \varphi _{\rm {nil}}^{s}}$

for a fixed ${\displaystyle {}s<k}$, and a polynomial ${\displaystyle {}p(n)}$. The diagonal entries of ${\displaystyle {}p(n)\varphi _{\rm {diag}}^{n-s}}$ are (after diagonalization) of the form

${\displaystyle p(n)z_{j}^{n-s};}$

because of ${\displaystyle {}\vert {z_{j}}\vert <1}$, this converges for ${\displaystyle {}n\rightarrow \infty }$ to ${\displaystyle {}0}$. Therefore, ${\displaystyle {}p(n)\varphi _{\rm {diag}}^{n-s}}$ converges to the zero mapping, and this holds due to exercise also for the product with the fixed mapping ${\displaystyle {}\varphi _{\rm {nil}}^{s}}$. Hence, the sum converges to the zero mapping.