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From: HardySpicer on 15 Jun 2010 00:11
If I have a matrix A whose eigenvalues have all magnitude less than 1,
then does A^n approach zero for large integer n?


Hardy
From: TCL on 15 Jun 2010 00:30
On Jun 15, 12:11 am, HardySpicer <gyansor...(a)gmail.com> wrote:
> If I have a matrix A whose eigenvalues have all magnitude less than 1,
> then does A^n approach zero for large integer n?
>
> Hardy

Yes.
From: adamk on 14 Jun 2010 20:54
> If I have a matrix A whose eigenvalues have all
> magnitude less than 1,
> then does A^n approach zero for large integer n?
>
>
> Hardy

If t is an eigenvalue of A, what are the eigenvalues
of A^n.?
From: Arturo Magidin on 15 Jun 2010 01:13
On Jun 14, 11:11 pm, HardySpicer <gyansor...(a)gmail.com> wrote:
> If I have a matrix A whose eigenvalues have all magnitude less than 1,
> then does A^n approach zero for large integer n?

Not necessarily; take

( 0 -1 0 )
( 1 0 0 )
( 0 0 .5 )

The only eigenvalue is 1/2, A^n does not approach 0. The 2 x 2
submatrix corresponding to the upper left corner alternates between

( 0 -1 )
( 1 0 ),

( -1 0 )
( 0 -1 ),

( 0 1 )
(-1 0 ),

and

( 1 0 )
( 0 1 ),

so the matrix never "approaches 0".

If A is diagonalizable, then the answer is "yes" for suiitable notion
of "approach 0".

--
Arturo Magidin
From: adamk on 14 Jun 2010 21:51
> On Jun 14, 11:11 pm, HardySpicer
> <gyansor...(a)gmail.com> wrote:
> > If I have a matrix A whose eigenvalues have all
> magnitude less than 1,
> > then does A^n approach zero for large integer n?
>
> Not necessarily; take
>
> ( 0 -1 0 )
> ( 1 0 0 )
> ( 0 0 .5 )
>
> The only eigenvalue is 1/2, A^n does not approach 0.
> The 2 x 2
> submatrix corresponding to the upper left corner
> alternates between
>
> ( 0 -1 )
> ( 1 0 ),
>
> ( -1 0 )
> ( 0 -1 ),
>
> ( 0 1 )
> (-1 0 ),
>
> and
>
> ( 1 0 )
> ( 0 1 ),
>
> so the matrix never "approaches 0".
>
> If A is diagonalizable, then the answer is "yes" for
> suiitable notion
> of "approach 0".
>
> --
> Arturo Magidin

What then is wrong with the fact that if k is
an eigenvalue and v is a non-zero eigenvector:

A*v=k*v , then A^n*v=k^n*v , so that, for

|k|<1 , as n->oo, k^n->0, so that (some hand-waving)

A^n*v= k^n*v-> 0 , so that A^n*v=0 . ?
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