eigenvalues of powers of a matrix

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loftus@math.binghamton.edu - 23 Jun 2007 19:15 GMT

Hi guys,

Here's my question: Given a square matrix A, with
k eigenvalues e1, e2, ..., ek, and positive integer r,
are all of the eigenvalues of A^r in the set
(e1)^r, (e2)^r, ..., (ek)^r ?

That is, as you raise A to powers, can you ever pick up
new eigenvalues that are not powers of the eigenvalues
of A?
Thanks, John L.

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Helmut Jarausch - 23 Jun 2007 19:57 GMT

> Hi guys,
>
> Here's my question: Given a square matrix A, with
> k eigenvalues e1, e2, ..., ek, and positive integer r,
> are all of the eigenvalues of A^r in the set
> (e1)^r, (e2)^r, ..., (ek)^r ?

Yes

> That is, as you raise A to powers, can you ever pick up
> new eigenvalues that are not powers of the eigenvalues
> of A?

No.

If A*uk = uk*ek, than A*(A*uk) = uk*ek*ek, and so on.

Signature

Helmut Jarausch

Lehrstuhl fuer Numerische Mathematik
RWTH - Aachen University
D 52056 Aachen, Germany

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loftus@math.binghamton.edu - 23 Jun 2007 20:07 GMT

Hi Helmut. I think you've oversimplified the problem. That is, your
observation
below only shows that the set S := {(e1)^r, (e2)^r, ..., (ek)^r } is
contained in
the set T := all eigenvalues of A^r. But I'm trying to show the
converse, that is,
that T is a subset of S. (which would give that T = S)

Thanks for your help, John L.

> > Hi guys,
> >
[quoted text clipped - 19 lines]
> RWTH - Aachen University
> D 52056 Aachen, Germany

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Ioannis - 23 Jun 2007 20:38 GMT

Ο wrote:

> Hi guys,
>
[quoted text clipped - 6 lines]
> new eigenvalues that are not powers of the eigenvalues
> of A?

No. Assume otherwise. Call the eigenvalue e and the non-zero eigenvector it
corresponds to v_e.

The eigenvectors of A^n are fixed directionally for all n. This means that v_e
has the direction of some eigenvector v_{e_m}. In other words, there exists a
\in R and some m in N, such that:

v_e=a*v_{e_m} (1)

Consequently:

A^r*v_e=a*A^r*v_{e_m}=a*(e_m)^r*v_{e_m} (2)

But also from (1):

A^r*v_e=e*v_e=a*e*v_{e_m} (3)

Subtract (3) and (2) to get:

v_{e_m}*a*(e-(e_m)^r)=0 (4)

We assumed that the eigenvector v_{e_m}=/=0, so (4) implies:

a=0 or e=e_m^r, both of which are contradictions.

> Thanks, John L.

Signature

I.N. Galidakis --- http://ioannis.virtualcomposer2000.com/
----------------------------------------------------------
"There's ALWAYS a mistake somewhere"

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loftus@math.binghamton.edu - 23 Jun 2007 21:29 GMT

Hi Ioannis. I don't see why the eigenvectors of A^r must all have
the same "direction" as some in A. Also, I suppose you mean
the constant a to be a complex number, as the entries in the
eigenvectors can be imaginary, right?

Why must v_e be a scalar multiple of some v_{e_m} ?

Thanks for your input, John

> Ï wrote:
> > Hi guys,
[quoted text clipped - 38 lines]
> ----------------------------------------------------------
> "There's ALWAYS a mistake somewhere"

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Ioannis - 24 Jun 2007 00:15 GMT

Ο wrote:

> Hi Ioannis. I don't see why the eigenvectors of A^r must all have
> the same "direction" as some in A. Also, I suppose you mean
> the constant a to be a complex number, as the entries in the
> eigenvectors can be imaginary, right?

Yes, if you are talking about A's in C^n. I was talking about A's in R^n.

> Why must v_e be a scalar multiple of some v_{e_m} ?

Hint: Pick an eigenvector of A. Any one. Apply A onto it repeatedly. What
happens?

> Thanks for your input, John

Signature

I.N. Galidakis --- http://ioannis.virtualcomposer2000.com/
----------------------------------------------------------
"There's ALWAYS a mistake somewhere"

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loftus@math.binghamton.edu - 24 Jun 2007 00:48 GMT

Dear Ioannis,

The application of your "hint" only shows that if you apply A
repeatedly to an eigenvector v of A, you get an eigenvector of
A^r that is a scalar multiple of v. But it does not show that all of
the eigenvectors of A^r are scalar multiples of some eigenvector of
A.

Of course all of the eigenvectors of A are also eigenvectors of A^r,
with corresponding eigenvalues that are powers of those of A.

I believe there is a problem with your logic.

Thanks, John

> Ï wrote:
> > Hi Ioannis. I don't see why the eigenvectors of A^r must all have
[quoted text clipped - 14 lines]
> ----------------------------------------------------------
> "There's ALWAYS a mistake somewhere"

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Ioannis - 24 Jun 2007 02:13 GMT

Ο wrote:

> Dear Ioannis,
>
> The application of your "hint" only shows that if you apply A
> repeatedly to an eigenvector v of A, you get an eigenvector of
> A^r that is a scalar multiple of v.

Right.

> But it does not show that all of
> the eigenvectors of A^r are scalar multiples of some eigenvector of
> A.

I don't understand why you can't see it (or maybe I see it wrongly?) :-(

Pick any eigenvector e_v of A^r. Its direction must be _a fortiori_ one of the
fixed directions we have after A acts on our space exactly r times. There are
exactly n of those directions, where n is A's dimension. This implies that e_v
is _exactly_ a power r multiple of one of the n eigenvectors of A, since both
share the same direction.

Do you see a problem with the above?

> Of course all of the eigenvectors of A are also eigenvectors of A^r,
> with corresponding eigenvalues that are powers of those of A.
>
> I believe there is a problem with your logic.

Might very well be. Trust your instict. :-)

[snip]

Signature

I.N. Galidakis --- http://ioannis.virtualcomposer2000.com/
----------------------------------------------------------
"There's ALWAYS a mistake somewhere"

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Robert Israel - 24 Jun 2007 08:56 GMT

> Dear Ioannis,
>
[quoted text clipped - 3 lines]
> the eigenvectors of A^r are scalar multiples of some eigenvector of
> A=2E

They might not be. For example, consider the eigenvectors of A^2, where
[ 1 0 ] [ 0 1 ]
A = [ 0 -1 ] or [ 0 0 ]

> Of course all of the eigenvectors of A are also eigenvectors of A^r,
> with corresponding eigenvalues that are powers of those of A.

> I believe there is a problem with your logic.

However, IIRC the original question had to do with eigenvalues, not
eigenvectors. It is true that, for any positive integer r, the set of
eigenvalues of A^r is { t^r: t an eigenvalue of A}. In fact, for any
polynomial f (or even more generally, for any function analytic in a
neighbourhood of the spectrum of A) the eigenvalues of f(A) are
{f(r): r an eigenvalue of A}, and the algebraic multiplicity of
t as an eigenvalue of f(A) is the sum of the algebraic multiplicities
of r as eigenvalues of A for those r with f(r) = t.

Signature

Robert Israel israel@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada V6T 1Z2

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Math Forum / Mathematics / Numerical Analysis / June 2007

eigenvalues of powers of a matrix