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From: achille on 6 Jun 2010 09:58
On Jun 6, 7:49 pm, Ron <ron.sper...(a)gmail.com> wrote:
> 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.

It simply refers to the fact the two embedding S(0) and S(1) of the
sphere (to be precise, the n-dim ball which is an orientable n-
manifold)
of unit volume into R^n has different orientations. The orientation of
one of S(0) and S(1) will be opposite to the standard orientation of
R^n.
From: Ron on 7 Jun 2010 00:08
On Jun 6, 9:58 am, achille <achille_...(a)yahoo.com.hk> wrote:
> On Jun 6, 7:49 pm, Ron <ron.sper...(a)gmail.com> wrote:
>
>
>
>
>
> > 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.
>
> It simply refers to the fact the two embedding S(0) and S(1) of the
> sphere  (to be precise, the n-dim ball which is an orientable n-
> manifold)
> of unit volume into R^n has different orientations. The orientation of
> one of S(0) and S(1) will be opposite to the standard orientation of
> R^n.

This still doesn't say how you are defining volume here. I'm not
trying to knock down your argument since I see what you are trying to
do here, but what is your definition of volume? Absolutely A and B
give different orientations and I think I see what your argument is,
but I think if you are going to use "volume" in a sense that isn't
some kind of measure, then it needs to be defined.
From: Ron on 7 Jun 2010 00:15
On Jun 6, 9:58 am, achille <achille_...(a)yahoo.com.hk> wrote:
> On Jun 6, 7:49 pm, Ron <ron.sper...(a)gmail.com> wrote:
>
>
>
>
>
> > 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.
>
> It simply refers to the fact the two embedding S(0) and S(1) of the
> sphere  (to be precise, the n-dim ball which is an orientable n-
> manifold)
> of unit volume into R^n has different orientations. The orientation of
> one of S(0) and S(1) will be opposite to the standard orientation of
> R^n.

One can avoid the whole issue of "volume" of course by simply
considering the function f:I->R by f(t)=det((1-t)A+tB)
This is continuous and since either f(0)=1 and f(1)=-1 or f(0)=-1 and
f(1)=1, the intermediate value theorem says that there is a t with
f(t)=0. So you still get t with det((1-t)A+tB)=0 and the rest of the
argument follows just as yours did.
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