Algebra, Problems For Fun (17)

**NonCommAlg** · June 12th 2009, 04:56 PM

A square matrix $J$ with entries from $\mathbb{C}$ is called a Jordan matrix if $J=\begin{pmatrix}\lambda & 1 & 0 & 0 & \cdots & 0 \\ 0 & \lambda & 1 & 0 & \cdots & 0 \\ . & . & . & . & \cdots & . \\ . & . & . & . & \cdots & . \\ . & . & . & . & \cdots & . \\ 0 & 0 & 0 & 0 & \cdots & \lambda \end{pmatrix},$ for some $\lambda \in \mathbb{C}.$ Show that if $|\lambda| < 1,$ then $\lim_{n\to\infty}J^n = \bold{0}.$

Remark: Using Jordan normal form the above result is easily extended to any matrix $A$ with entries from $\mathbb{C}$ and with this property that $|\lambda| < 1,$ for any eigenvalue $\lambda$ of $A.$

**PaulRS** · June 12th 2009, 06:18 PM

We write: $ J = \lambda \cdot {\text{Id}} + M $ where $ M = \left( {\begin{array}{*{20}c} 0 & 1 & 0 & \cdots & 0 \\ {} & 0 & {} & {} & \vdots \\ \vdots & {} & \ddots & {} & 0 \\ {} & {} & {} & 0 & 1 \\ 0 & {} & \cdots & {} & 0 \\ \end{array} } \right) $

Now note that M represents the following digraph: 1->2->3->...->n . When we exponentiate M (remember that $ \left[ {M^m } \right]_{i,j} $ is the number of paths of lenght m from i to j ) we get: $ \left[ {M^m } \right]_{i,j} = \left\{ \begin{gathered} 1{\text{ if }}i + m \leqslant {\text{dimension and }}j = i + m \hfill \\ 0{\text{ otherwise}} \hfill \\ \end{gathered} \right. $ (1) (when I say dimension I am talking about the dimension of our matrices) - From here it is clear that $ M^{{\text{dimension}}} =\bold{0} $ . (2)

Where does this come in handy? Well: $ J^n = \left( {\lambda \cdot {\text{Id}} + M} \right)^n = \sum\limits_{k = 0}^n {\binom{n}{k}\cdot M^k \cdot \lambda ^{n - k} } $ ( $M^0=\text{Id}$ )

Consider $n\geq \text{dimension}-1$ then - by (2) - $ J^n = \sum\limits_{k = 0}^{{\text{dimension}-1}} {\binom{n}{k}\cdot M^k \cdot \lambda ^{n - k} } $

Then $ \left[ {J^n } \right]_{i,j} = \sum\limits_{k = 0}^{{\text{dimension}-1}} {\binom{n}{k}\left[ {M^k } \right]_{i,j} \cdot \lambda ^{n - k} } $ but, look back at (1)

if our entry here is not 0 already, it is only one of the terms of the sum , i.e. $ \left[ {J^n } \right]_{i,j} = \binom{n}{l} \lambda ^{n - l} $ for some $0\leq{l}\leq{\text{dimension}-1}$ and this clearly tends to 0 since $l$ is fixed ; $\binom{n}{l}\approx{\tfrac{n^l}{l!}}$ and $|\lambda|<1$

**NonCommAlg** · June 12th 2009, 07:56 PM

Originally Posted by PaulRS

We write: $ J = \lambda \cdot {\text{Id}} + M $ where $ M = \left( {\begin{array}{*{20}c} 0 & 1 & 0 & \cdots & 0 \\ {} & 0 & {} & {} & \vdots \\ \vdots & {} & \ddots & {} & 0 \\ {} & {} & {} & 0 & 1 \\ 0 & {} & \cdots & {} & 0 \\ \end{array} } \right) $

Now note that M represents the following digraph: 1->2->3->...->n . When we exponentiate M (remember that $ \left[ {M^m } \right]_{i,j} $ is the number of paths of lenght m from i to j ) we get: $ \left[ {M^m } \right]_{i,j} = \left\{ \begin{gathered} 1{\text{ if }}i + m \leqslant {\text{dimension and }}j = i + m \hfill \\ 0{\text{ otherwise}} \hfill \\ \end{gathered} \right. $ (1) (when I say dimension I am talking about the dimension of our matrices) - From here it is clear that $ M^{{\text{dimension}}} =\bold{0} $ . (2)

Where does this come in handy? Well: $ J^n = \left( {\lambda \cdot {\text{Id}} + M} \right)^n = \sum\limits_{k = 0}^n {\binom{n}{k}\cdot M^k \cdot \lambda ^{n - k} } $ ( $M^0=\text{Id}$ )

Consider $n\geq \text{dimension}-1$ then - by (2) - $ J^n = \sum\limits_{k = 0}^{{\text{dimension}-1}} {\binom{n}{k}\cdot M^k \cdot \lambda ^{n - k} } $

Then $ \left[ {J^n } \right]_{i,j} = \sum\limits_{k = 0}^{{\text{dimension}-1}} {\binom{n}{k}\left[ {M^k } \right]_{i,j} \cdot \lambda ^{n - k} } $ but, look back at (1)

if our entry here is not 0 already, it is only one of the terms of the sum , i.e. $ \left[ {J^n } \right]_{i,j} = \binom{n}{l} \lambda ^{n - l} $ for some $0\leq{l}\leq{\text{dimension}-1}$ and this clearly tends to 0 since $l$ is fixed ; $\binom{n}{l}\approx{\tfrac{n^l}{l!}}$ and $|\lambda|<1$

beautiful! now suppose that $A$ is a square matrix with entries in $\mathbb{C}.$ let $\lambda_1, \cdots , \lambda_r$ be all distinct eigenvalues of $A.$ then we have: $A=P^{-1}BP,$ where $P$ is some invertible matrix and

$B=\begin{pmatrix}J_1 & & & & & & \\ & J_2 & & & & & \\ & & . & & & & \\ & & & . & & & \\ & & & & . & & \\ & & & & & & J_r \end{pmatrix},$ where each $J_k$ is a Jordan matrix with $\lambda_k$ on the diagonal. anywhere else in $B$ is 0. therefore: $A^n=P^{-1}B^nP=P^{-1}\begin{pmatrix}J_1^n & & & & & & \\ & J_2^n & & & & & \\ & & . & & & & \\ & & & . & & & \\ & & & & . & & \\ & & & & & & J_r^n \end{pmatrix}P$

and thus if $|\lambda_k|<1$ for all $k,$ then $\lim_{n\to\infty}A^n=\bold{0}.$ this is an important result in applied and pure linear algebra and a key to the solution of skeboy's question in http://www.mathhelpforum.com/math-he...-matrices.html.

skeboy doesn't seem to be looking for a rigorous proof of his question, which, by the way, is a part of the beautiful Perron-Frobenius theorem. i'll post my solution to his problem later.

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