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Math Help - Algebra, Problems For Fun (17)

  1. #1
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    Algebra, Problems For Fun (17)

    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.
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  2. #2
    Super Member PaulRS's Avatar
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    We write: <br />
J = \lambda  \cdot {\text{Id}} + M<br />
where <br />
M = \left( {\begin{array}{*{20}c}<br />
   0 & 1 & 0 &  \cdots  & 0  \\<br />
   {} & 0 & {} & {} &  \vdots   \\<br />
    \vdots  & {} &  \ddots  & {} & 0  \\<br />
   {} & {} & {} & 0 & 1  \\<br />
   0 & {} &  \cdots  & {} & 0  \\<br /> <br />
 \end{array} } \right)<br />

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

    Where does this come in handy? Well: <br />
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} } <br />
( M^0=\text{Id} )

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

    Then <br />
\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} } <br />
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. <br />
\left[ {J^n } \right]_{i,j}  = \binom{n}{l} \lambda ^{n - l} <br />
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
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  3. #3
    MHF Contributor

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    Quote Originally Posted by PaulRS View Post
    We write: <br />
J = \lambda \cdot {\text{Id}} + M<br />
where <br />
M = \left( {\begin{array}{*{20}c}<br />
0 & 1 & 0 & \cdots & 0 \\<br />
{} & 0 & {} & {} & \vdots \\<br />
\vdots & {} & \ddots & {} & 0 \\<br />
{} & {} & {} & 0 & 1 \\<br />
0 & {} & \cdots & {} & 0 \\<br /> <br />
\end{array} } \right)<br />

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

    Where does this come in handy? Well: <br />
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} } <br />
( M^0=\text{Id} )

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

    Then <br />
\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} } <br />
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. <br />
\left[ {J^n } \right]_{i,j} = \binom{n}{l} \lambda ^{n - l} <br />
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<br />
\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<br />
\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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