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From: Ron on 5 Jun 2010 08:05
On Jun 5, 7:23 am, Kaba <n...(a)here.com> wrote:
> 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?
>

I suspect the idea is this. A is invertible so A+B is singular if and
only if I+A^{-1}B is singular. So we simply consider the case where
A=I and B is an orthogonal matrix of determinant -1. The argument is
essentially the same as the earlier one using the fact that an
orthogonal matrix of determinant -1 has -1 as an eigenvalue.
> --http://kaba.hilvi.org

From: Kaba on 5 Jun 2010 08:34
Ron wrote:
> On Jun 5, 7:23 am, Kaba <n...(a)here.com> wrote:
> > 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?
> >
>
> I suspect the idea is this. A is invertible so A+B is singular if and
> only if I+A^{-1}B is singular. So we simply consider the case where
> A=I and B is an orthogonal matrix of determinant -1. The argument is
> essentially the same as the earlier one using the fact that an
> orthogonal matrix of determinant -1 has -1 as an eigenvalue.

Ok, I see, it works:)

--
http://kaba.hilvi.org
From: achille on 5 Jun 2010 08:39
On Jun 5, 4:10 am, Kaba <n...(a)here.com> wrote:
> 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
How about this proof:


Let S be the unit sphere and for 0 <= t <= 1, S(t) be
S's image under the linear map x -> ((1-t) A + t B) x.
The volume of S(0) is det(A), the volume of S(1) is det(B).
Since det(A) = -det(B), there is a 0 < t < 1 such that
volume of S(t) = 0. ie. det( (1-t)A + tB ) = 0.
For such t, there is a non-zero vector y such that
((1-t)A + tB) y = 0
=> (1-t)A y = -tB y,
=> (1-t) |y| = (1-t)|Ay| = t |B y| = t |y|
=> 1-t = t
=> t = 1/2
=> det((A+B)/2) = 0.


From: Kaba on 5 Jun 2010 09:09
achille wrote:
> How about this proof:
>
>
> Let S be the unit sphere and for 0 <= t <= 1, S(t) be

Here you probably mean a ball with unit volume rather than with unit
radius.

> S's image under the linear map x -> ((1-t) A + t B) x.
> The volume of S(0) is det(A), the volume of S(1) is det(B).
> Since det(A) = -det(B), there is a 0 < t < 1 such that
> volume of S(t) = 0. ie. det( (1-t)A + tB ) = 0.
> For such t, there is a non-zero vector y such that
> ((1-t)A + tB) y = 0
> => (1-t)A y = -tB y,
> => (1-t) |y| = (1-t)|Ay| = t |B y| = t |y|
> => 1-t = t
> => t = 1/2
> => det((A+B)/2) = 0.

You are very creative at coming up with these proofs. Certainly more
geometrical. It is interesting that t happens to be 1/2.

--
http://kaba.hilvi.org
From: Ron on 6 Jun 2010 07:49
On Jun 5, 8:39 am, achille <achille_...(a)yahoo.com.hk> wrote:
> On Jun 5, 4:10 am, Kaba <n...(a)here.com> wrote:
>
>
>
> > 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
>
> How about this proof:
>
> Let S be the unit sphere and for 0 <= t <= 1, S(t) be
> S's image under the linear map x -> ((1-t) A + t B) x.
> The volume of S(0) is det(A), the volume of S(1) is det(B).
> Since det(A) = -det(B), there is a 0 < t < 1 such that
> volume of S(t) = 0. ie. det( (1-t)A + tB ) = 0.
> For such t, there is a non-zero vector y such that
>     ((1-t)A + tB) y = 0
> =>   (1-t)A y = -tB y,
> =>  (1-t) |y| = (1-t)|Ay| = t |B y| = t |y|
> => 1-t = t
> => t = 1/2
> => det((A+B)/2) = 0.

I'm not sure what you mean by "volume" here as volume as generally
defined is a positive number and you are arguing that either S(0) or
S(1) has negative volume.
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