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From: Kaba on 29 Apr 2010 07:54
Hi,

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?

--
http://kaba.hilvi.org
From: Robert Israel on 29 Apr 2010 11:21
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?

Hint: What kind of matrix is iA?
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
From: Stephen Montgomery-Smith on 29 Apr 2010 12:17
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?
>
> Hint: What kind of matrix is iA?

Or a hint in a slightly different direction: what is the value of x^T A
x for any vector x?
From: Kaba on 29 Apr 2010 13:20
Stephen Montgomery-Smith wrote:
> 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?
> >
> > Hint: What kind of matrix is iA?
>
> Or a hint in a slightly different direction: what is the value of x^T A
> x for any vector x?

for all x:
x^T A x = 0.5 x^T A x - 0.5 x^T A^T x = 0.5 x^T A x - 0.5 x^T A x = 0

Considering an eigenvalue m of A:
Ax = mx
=>
for all x: x^T A x = m x^T x = 0
=>
m = 0

Eigenvalues of A are all the numbers m such that det(A - mI) = 0.
Therefore:
m != 0 <=> det(A - mI) != 0 <=> (A - mI) is invertible.

Substituting m = -1 shows A + I is invertible.

Thanks!

How about the iA hint? All I got was that iA is Hermitian.

--
http://kaba.hilvi.org
From: Robert Israel on 29 Apr 2010 18:56
Kaba <none(a)here.com> writes:

> Stephen Montgomery-Smith wrote:
> > 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?
> > >
> > > Hint: What kind of matrix is iA?
> >
> > Or a hint in a slightly different direction: what is the value of x^T A
> > x for any vector x?
>
> for all x:
> x^T A x = 0.5 x^T A x - 0.5 x^T A^T x = 0.5 x^T A x - 0.5 x^T A x = 0
>
> Considering an eigenvalue m of A:
> Ax = mx
> =>
> for all x: x^T A x = m x^T x = 0
> =>
> m = 0
>
> Eigenvalues of A are all the numbers m such that det(A - mI) = 0.
> Therefore:
> m != 0 <=> det(A - mI) != 0 <=> (A - mI) is invertible.
>
> Substituting m = -1 shows A + I is invertible.
>
> Thanks!
>
> How about the iA hint? All I got was that iA is Hermitian.

And what do you know about eigenvalues of Hermitian matrices?
--
Robert Israel israel(a)math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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