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From: Maarten Bergvelt on 4 Jun 2010 11:14
On 2010-06-04, Kaba <none(a)here.com> wrote:
> Maarten Bergvelt wrote:
>> On 2010-06-04, Kaba <none(a)here.com> wrote:
>> > Hi,
>> >
>> > This is part of the first question for chapter 1 of "Applied numerical
>> > linear algebra" book, but I just can't come up with a solution:
>> >
>> > If A and B are orthogonal matrices, and det(A) = -det(B), show that
>> > A + B is singular.
>> >
>> > Any hints?
>> >
>> Hm, good question. What about looking for a null vector of A+B? Or
>> showing that A^tB has eigenvalue -1? Not sure if that helps.
>
> Thanks, I'll look into those.

Maybe try first a special case. We know that det(A)=+/-1, so that A
could be the identity, and then B must have det -1. The your statement says
that
(1+B)x=0, for x=/ 0,
so that B has an eigenvector with eigenvalue -1, in case B
is improper orthogonal. Is this always true?


> Meanwhile, I got a partial result:
>
> C = (A + B)^T (A - B)
>= A^T A - B^T B + B^T A - A^T B
>= B^T A - A^T B
>
> C^T = A^T B - B^T A = -C
>=>
> det(C) = 0
>=>
> det(A + B)det(A - B) = 0
>=>
> det(A + B) = 0 or det(A - B) = 0
>
> But that's not enough. The additional information det(A) = -det(B) was
> not used here.
>


--
Maarten Bergvelt
From: Kaba on 4 Jun 2010 11:44
> On 2010-06-04, Kaba <none(a)here.com> wrote:
> >> > Hi,
> >> >
> >> > This is part of the first question for chapter 1 of "Applied numerical
> >> > linear algebra" book, but I just can't come up with a solution:
> >> >
> >> > If A and B are orthogonal matrices, and det(A) = -det(B), show that
> >> > A + B is singular.
> >> >
> >> > Any hints?

Maarten Bergvelt wrote:

> Maybe try first a special case. We know that det(A)=+/-1, so that A
> could be the identity, and then B must have det -1. The your statement says
> that
> (1+B)x=0, for x=/ 0,
> so that B has an eigenvector with eigenvalue -1, in case B
> is improper orthogonal. Is this always true?

Aha! Let x != 0.

First we have:

(A + B)x = 0
<=>
Ax = -Bx
<=>
x = -A^T Bx
<=>
(I + A^T B)x = 0
<=>
A^T B x = -x
<=>
A^T B has an eigenvalue -1.

Then we also have:

det(A) = -det(B)
<=>
det(A) det(B) = -1
<=>
det(A^T B) = -1

A^T B is orthogonal and an orthogonal matrix has eigenvalues either -1
or +1. Therefore A^T B has at least one eigenvalue -1. Thus there is a
non-zero null vector for (A + B) and (A + B) is singular.

Thanks.

--
http://kaba.hilvi.org
From: Maarten Bergvelt on 4 Jun 2010 14:08
On 2010-06-04, Kaba <none(a)here.com> wrote:
>> On 2010-06-04, Kaba <none(a)here.com> wrote:
>> >> > Hi,
>> >> >
>> >> > This is part of the first question for chapter 1 of "Applied numerical
>> >> > linear algebra" book, but I just can't come up with a solution:
>> >> >
>> >> > If A and B are orthogonal matrices, and det(A) = -det(B), show that
>> >> > A + B is singular.
>> >> >
>> >> > Any hints?
>
> Maarten Bergvelt wrote:
>
>> Maybe try first a special case. We know that det(A)=+/-1, so that A
>> could be the identity, and then B must have det -1. The your statement says
>> that
>> (1+B)x=0, for x=/ 0,
>> so that B has an eigenvector with eigenvalue -1, in case B
>> is improper orthogonal. Is this always true?
>
> Aha! Let x != 0.
>
> First we have:
>
> (A + B)x = 0
><=>
> Ax = -Bx
><=>
> x = -A^T Bx
><=>
> (I + A^T B)x = 0
><=>
> A^T B x = -x
><=>
> A^T B has an eigenvalue -1.
>
> Then we also have:
>
> det(A) = -det(B)
><=>
> det(A) det(B) = -1
><=>
> det(A^T B) = -1
>
> A^T B is orthogonal and an orthogonal matrix has eigenvalues either -1
> or +1. Therefore A^T B has at least one eigenvalue -1. Thus there is a
> non-zero null vector for (A + B) and (A + B) is singular.

Uh, not all orthogonal matrices have eigenvalues +1 or -1, for
instance a rotation matrix has no real eigenvalues at all, if the
angle of rotation does not happen to be special. In general the
eigenvalues will be complex numbers of norm 1.

So what we are saying is that if C is a orthogonal matrix of det -1
then it has a -1 eigenvalue.

I would think about decomposing C as a product of a rotation and a
reflection.



--
Maarten Bergvelt
From: achille on 4 Jun 2010 15:19
On Jun 4, 9:48 pm, Kaba <n...(a)here.com> wrote:
> Hi,
>
> This is part of the first question for chapter 1 of "Applied numerical
> linear algebra" book, but I just can't come up with a solution:
>
> If A and B are orthogonal matrices, and det(A) = -det(B), show that
> A + B is singular.
>
> Any hints?
>
> --http://kaba.hilvi.org

Hint: A^t (A+B) B^t = B^t + A^t and take det(.) on both sides.
From: Kaba on 4 Jun 2010 15:57
Maarten Bergvelt wrote:
> > A^T B is orthogonal and an orthogonal matrix has eigenvalues either -1
> > or +1. Therefore A^T B has at least one eigenvalue -1. Thus there is a
> > non-zero null vector for (A + B) and (A + B) is singular.
>
> Uh, not all orthogonal matrices have eigenvalues +1 or -1, for
> instance a rotation matrix has no real eigenvalues at all, if the
> angle of rotation does not happen to be special. In general the
> eigenvalues will be complex numbers of norm 1.

Right:/

> So what we are saying is that if C is a orthogonal matrix of det -1
> then it has a -1 eigenvalue.

Yep.

> I would think about decomposing C as a product of a rotation and a
> reflection.

Hmm..

--
http://kaba.hilvi.org
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