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From: JEMebius on 1 May 2010 19:04
Kaba wrote:
> 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.
>

Study hints:
(A) make a study of Robert Israel's hint.
(B) A skew-symmetric real matrix is a skew-Hermitian complex matrix.
(C) Let A be skew-Hermitian. Prove that all eigenvalues of I + A have real part 1.

(D)
Extend your explorations and studies to the Cayley Transform (See Wikipedia
http://en.wikipedia.org/wiki/Cayley_transform)
From: Kaba on 1 May 2010 19:34
JEMebius wrote:
> > How about the iA hint? All I got was that iA is Hermitian.
> >
>
> Study hints:
> (A) make a study of Robert Israel's hint.
> (B) A skew-symmetric real matrix is a skew-Hermitian complex matrix.
> (C) Let A be skew-Hermitian. Prove that all eigenvalues of I + A have real part 1.

All done in reply to Robert Israel:)

> (D)
> Extend your explorations and studies to the Cayley Transform (See Wikipedia
> http://en.wikipedia.org/wiki/Cayley_transform)

Heh, I actually started this thread because of reading from that page
the sentence "Then I + A is invertible.. ". I could almost write "what a
coincidence", but probably it is not.

--
http://kaba.hilvi.org
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