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From: Kaba on 4 Jun 2010 16:10
achille wrote:
> 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.

Well, that's straightforward, thanks:)

I am not sure if this is a best kind of exercise, at least when
separated from any context. It seems this is just a trick, with no
deeper lesson to learn. But maybe it is used somewhere in the next
pages.

--
http://kaba.hilvi.org
From: Stephen Montgomery-Smith on 5 Jun 2010 01:10
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?
>

Here is my attempt. If A is orthogonal, then the determinant is either
1 or minus 1. WLOG det(A) = 1. Now

A+B = A(I + A^{-1} B)

so WLOG A=I, and B is an orthogonal matrix whose determinant is -1.

The eigenvalues of B are complex numbers with absolute value 1. The
complex roots come on conjugate pairs, so the product of those is 1.
Therefore B must have at least one eigenvalue equal to -1. Hence I+B is
singular.


From: Ostap Bender on 5 Jun 2010 01:22
On Jun 4, 6:48 am, 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.

Look up the theorem that says that a square matrix X is singular iff
det(X) = 0.

But det (A + B) = det A + det B = - det(B) + det(B) = 0

From: Ostap Bender on 5 Jun 2010 01:26
On Jun 4, 10:22 pm, Ostap Bender <ostap_bender_1...(a)hotmail.com>
wrote:
> On Jun 4, 6:48 am, 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.
>
> Look up the theorem that says that a square matrix X is singular iff
> det(X) = 0.
>
> But det (A + B) = det A + det B =  - det(B) +  det(B) = 0

Never mind: determinant is not a linear function. Sorry.
From: Kaba on 5 Jun 2010 07:23
Stephen Montgomery-Smith wrote:
> 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?
> >
>
> Here is my attempt. If A is orthogonal, then the determinant is either
> 1 or minus 1. WLOG det(A) = 1. Now
>
> A+B = A(I + A^{-1} B)
>
> so WLOG A=I, and B is an orthogonal matrix whose determinant is -1.

Could you be more specific why WLOG here?

--
http://kaba.hilvi.org
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