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From: Kaba on 1 May 2010 15:10
Stephen Montgomery-Smith wrote:
> If the eigenvalue is complex, then the eigenvector is complex as well.
> So you cannot conclude that x^T x is non-zero, which is an essential
> part of your argument.

Yes, I noticed that.

> You really need to then consider x^* A x, where * represents the
> conjugate transpose.

To quote myself (^H = conjugate transpose):

"Let's generalize to skew-Hermitian matrices. Then it holds that
for all x: x^H A x = 0

Then for eigenvector q and eigenvalue m:
Aq = mq
=>
q^H Aq = m q^H q = 0
=>
m = 0

Here q != 0, by the definition of an eigenvector, and thus q^H q > 0.
But this is wrong too, counter-example being given by:

A = [0, i]
[i, 0]

which is skew-Hermitian and has eigenvalues
m_1 = i
m_2 = -i
and eigenvectors
q_1 = [1, 1]
q_2 = [1, -1]

Where does the logic go wrong?"

--
http://kaba.hilvi.org
From: Stephen Montgomery-Smith on 1 May 2010 17:05
Kaba wrote:
> Stephen Montgomery-Smith wrote:
>> If the eigenvalue is complex, then the eigenvector is complex as well.
>> So you cannot conclude that x^T x is non-zero, which is an essential
>> part of your argument.
>
> Yes, I noticed that.
>
>> You really need to then consider x^* A x, where * represents the
>> conjugate transpose.
>
> To quote myself (^H = conjugate transpose):
>
> "Let's generalize to skew-Hermitian matrices. Then it holds that
> for all x: x^H A x = 0

No. For a general complex vector x, we can only say that x^H A x is
purely imaginary.
From: Kaba on 1 May 2010 17:31
Stephen Montgomery-Smith wrote:
> > "Let's generalize to skew-Hermitian matrices. Then it holds that
> > for all x: x^H A x = 0
>
> No. For a general complex vector x, we can only say that x^H A x is
> purely imaginary.

Ah, that's right:
x^H A x = -x^H A^H x != -(x^H A^H x)^H = -x^H A x

Thanks.

--
http://kaba.hilvi.org
From: Robert Israel on 2 May 2010 17:45
Kaba <none(a)here.com> writes:

> Robert Israel wrote:
> > Kaba <none(a)here.com> writes:
> > > > >> Problem:
> > > > >> Let A be a skew-symmetric matrix over the reals, i.e. A^T = -A.
> > > > >> Show
> > > > >> that I + A is invertible.
> > > > >>
> > > > >> Any hints for a proof?
> >
> > > How about the iA hint? All I got was that iA is Hermitian.
> >
> > And what do you know about eigenvalues of Hermitian matrices?
>
> Eigenvalues of Hermitian matrices are real. A Hermitian matrix
> diagonalizes, thus:
>
> iA = U D U^H
>
> for some unitary U and real diagonal D. Thus:
>
> A = U (-i D) U^H
> =>
> A + aI = U (-iD + aI) U^H
>
> for all real a != 0.
>
> Therefore all eigenvalues of A + aI are non-zero. Because the
> determinant is the product of eigenvalues, det(A + aI) != 0.
> Thus A + aI is invertible. By substituting a = 1, we see that A + I is
> invertible.

Simpler: r is an eigenvalue of A iff ir is an eigenvalue of iA. So
eigenvalues of iA are real iff eigenvalues of A are imaginary.
And -1 is not imaginary.
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Kaba on 2 May 2010 17:57
Robert Israel wrote:
> > Therefore all eigenvalues of A + aI are non-zero. Because the
> > determinant is the product of eigenvalues, det(A + aI) != 0.
> > Thus A + aI is invertible. By substituting a = 1, we see that A + I is
> > invertible.
>
> Simpler: r is an eigenvalue of A iff ir is an eigenvalue of iA. So
> eigenvalues of iA are real iff eigenvalues of A are imaginary.
> And -1 is not imaginary.

True, thanks.

--
http://kaba.hilvi.org
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