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From: Rob Johnson on 15 Jun 2010 07:21
In article <98a219af-8682-44b3-bb6c-c2712befb9fc(a)q12g2000yqj.googlegroups.com>,
Arturo Magidin <magidin(a)member.ams.org> wrote:
>On Jun 14, 11:11=A0pm, 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".

The eigenvalues of your matrix are { i, -i, 1/2 }; two of the
eigenvalues have magnitude 1 or greater. I don't see how it
satisfies the OP's conditions.

Rob Johnson <rob(a)trash.whim.org>
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From: TCL on 15 Jun 2010 08:31
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

In fact,A^n approaches zero iff all eigenvalues of A have magnitude
less than 1.
From: Ray Vickson on 15 Jun 2010 11:40
On Jun 14, 10:13 pm, Arturo Magidin <magi...(a)member.ams.org> wrote:
> 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 )

This matrix violates the OP's assumption: its eigenvalues are 1/2, i
and -i (i = sqrt(-1)), so it has two eigenvalues of magnitude 1.

R.G. Vickson

>
> 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: Ray Vickson on 15 Jun 2010 12:48
On Jun 14, 9:11 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?

Yes. This follows from the Jordan Canonical Form; see
http://mathworld.wolfram.com/JordanCanonicalForm.html and
http://mathworld.wolfram.com/JordanMatrixDecomposition.html. There is
a matrix S such that A = S^(-1)*J*S, where J = Jordan canonical form
of A: J = diag(J1,J2,...,Jr), where the m_i x m_i Jordan block Ji has
Ji(u,v) = lambda_i for 1 <= u = v <= m_i (lambda_i is an eigenvalue of
mutiplicity at least m_i), Ji(u,u+1) = 1 for 1 <= u <= m_i-1 and all
other J(u,v) = 0; that is, Ji is upper-triangular with lambda_i on the
diagonal, 1 just above the diagonal and zero elsewhere. It follows
that there exist matrices Eik such that A^n = sum_{i=1..r} sum{k =
0..m_i-1} Eik * lambda_i^(n-k). If all |lambda_i| < 1 we have A^n -->
0 as n --> infinity.

R.G. Vickson

>
> Hardy

From: Ronald Bruck on 15 Jun 2010 12:52
In article
<98a219af-8682-44b3-bb6c-c2712befb9fc(a)q12g2000yqj.googlegroups.com>,
Arturo Magidin <magidin(a)member.ams.org> wrote:

> 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".

But this matrix has two eigenvalues of absolute value 1, so does not
satisfy the conditions.

The answer to the OP is "yes". The spectral radius r_\sigma(T) is
lim_n ||T^n||^(1/n), and is less than 1; therefore T^n --> 0.

-- Ron Bruck
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