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From: Kaba on 4 Jun 2010 09:48
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
From: Kaba on 4 Jun 2010 10:29
Kaba 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?

Hmm.. It seems I found one way:

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
=>
C = 0

Assume (A + B) is invertible. Then
A - B = 0
<=>
A = B
<=>
false, since det(A) != det(B).

Therefore (A + B) is singular. Of course, since A and B are orthogonal,
det(A) != det(B) <=> det(A) = -det(B).

Feels a bit arbitrary though... Can you see other ways?

--
http://kaba.hilvi.org
From: Kaba on 4 Jun 2010 10:33
Kaba wrote:
> C^T = A^T B - B^T A = -C
> =>
> C = 0

Oops, this does not hold.

--
http://kaba.hilvi.org
From: Maarten Bergvelt on 4 Jun 2010 10:52
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.

--
Maarten Bergvelt
From: Kaba on 4 Jun 2010 11:07
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. 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.

--
http://kaba.hilvi.org
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