Linear algebra (Osnabrück 2024-2025)/Part II/Lecture 53

Let ${\displaystyle {}w}$ denote the limit vector. Then

${\displaystyle {}\varphi ^{n+1}(v)=\varphi (\varphi ^{n}(v))\,}$

holds for all ${\displaystyle {}n\in \mathbb {N} }$. Because of the continuity (according to Theorem 52.17 ) of ${\displaystyle {}\varphi }$, the limit and the mapping may be swapped (according to Theorem 52.13 ). That is,

${\displaystyle {}w=\lim _{n\rightarrow \infty }\varphi ^{n}(v)=\lim _{n\rightarrow \infty }\varphi ^{n+1}(v)=\lim _{n\rightarrow \infty }\varphi (\varphi ^{n}(v))=\varphi {\left(\lim _{n\rightarrow \infty }\varphi ^{n}(v)\right)}=\varphi (w)\,.}$

Therefore, ${\displaystyle {}w}$ is a fixed point of ${\displaystyle {}\varphi }$; that means, it is the zero vector, or an eigenvector for the eigenvalue ${\displaystyle {}1}$.

This follows directly from

${\displaystyle {}\varphi ^{n}={\left(\varphi _{\rm {diag}}+\varphi _{\rm {nil}}\right)}^{n}\,,}$

the commuting property, and the general binomial formula.

(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 Lemma 53.4 . 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 28.13 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 53.7 also for the product with the fixed mapping ${\displaystyle {}\varphi _{\rm {nil}}^{s}}$. Hence, the sum converges to the zero mapping.

Let ${\displaystyle {}\lambda }$ be an eigenvalue of ${\displaystyle {}\varphi }$ with

${\displaystyle {}\vert {\lambda }\vert =\rho (\varphi )\,.}$

Let ${\displaystyle {}v\in V}$ be an eigenvector of ${\displaystyle {}\lambda }$. Then

${\displaystyle {}\vert {\lambda }\vert \cdot \Vert {v}\Vert =\Vert {\lambda v}\Vert =\Vert {\varphi (v)}\Vert \leq \Vert {\varphi }\Vert _{\rm {max}}\cdot \Vert {v}\Vert \,,}$

and division by ${\displaystyle {}\Vert {v}\Vert }$ yields the claim.

(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

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

${\displaystyle {}\varphi ^{n}(v)}$ is bounded. From (2) to (3) is clear. If (3) is fulfilled, and

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

is a linear combination, then

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

The boundedness of the sequences ${\displaystyle {}\varphi ^{n}(v_{j})}$ implies the boundedness of this sum sequence. The equivalence between (4) and (5) is clear, because over ${\displaystyle {}\mathbb {C} }$ the Jordan normal form exists, and the eigenvalues and their multiplicities can be read off from the Jordan blocks. From (2) to (5). We may assume ${\displaystyle {}{\mathbb {K} }=\mathbb {C} }$. Let

${\displaystyle {}J={\begin{pmatrix}\lambda &1&0&\cdots &0\\0&\lambda &1&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\\vdots &\cdots &0&\lambda &1\\0&\cdots &\cdots &0&\lambda \end{pmatrix}}\,}$

be a Jordan block of the Jordan normal form. In case

${\displaystyle {}\vert {\lambda }\vert >1\,,}$

for a corresponding eigenvector ${\displaystyle {}v}$ we obtain

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

directly contradicting boundedness. So let

${\displaystyle {}\vert {\lambda }\vert =1\,,}$

and assume that the length of the Jordan block is at least two. Due to Exercise 53.21 , we have

${\displaystyle {}{\begin{pmatrix}\lambda &1&0&\cdots &0\\0&\lambda &1&\cdots &0\\\vdots &\ddots &\ddots &\ddots &\vdots \\\vdots &\cdots &0&\lambda &1\\0&\cdots &\cdots &0&\lambda \end{pmatrix}}^{n}{\begin{pmatrix}1\\1\\0\\\vdots \\0\end{pmatrix}}={\begin{pmatrix}\lambda ^{n}+n\lambda ^{n-1}\\\lambda ^{n}\\0\\\vdots \\0\end{pmatrix}}\,.}$

Here, the first component is

${\displaystyle {}\lambda ^{n}+n\lambda ^{n-1}=\lambda ^{n-1}(\lambda +n)\,,}$

which is not bounded contradicting the condition.

To conclude from (5) to (1), we may consider each Jordan block separately, because, according to Exercise 53.19 , stability is compatible with a direct sum decomposition. For the first type, the statement follows from Theorem 53.6 . For the type ${\displaystyle {}\lambda }$ with ${\displaystyle {}\lambda =1}$, the statement is clear, because the norms of the powers equal ${\displaystyle {}1}$.