Re: Gram-Schmidt process

> On Sun, 27 Nov 2005 17:44:18 -0500, quasi
> <quasi@xxxxxxxx> wrote:
>
> >On Sun, 27 Nov 2005 17:32:07 -0500, quasi
> <quasi@xxxxxxxx> wrote:
> >
> >>On Sun, 27 Nov 2005 17:15:55 EST, SusanP
> <susanp@xxxxxxxxxxx> wrote:
> >>
> >>>Is it true that if {w1, w2, ..., wn} is an
> orthogonal
> >>>set of nonzero vectors, then the vectors v1,
> v2,..., vn
> >>>derived from the Gram-Schmidt process satisfy
> vi=wi,
> >>>for i= 1,2, ..., n?
> >>>
> >>>If so, can anyone think of a way to prove it?
> (Possibly
> >>>without induction... I don't like it very
> much....)
> >>>
> >>>I know the orthogonal vectors w1, w2,..., wn are
> >>>linearly independent.
> >>
> >>First, prove it for n=1.
> >>
> >>Then try it for n=2.
> >>
> >>Then n=3.
> >>
> >>Continue for a while ..., n=4, n=5, ...
> >>
> >>By the time you get to 100 and realize that you're
> still short of your
> >>goal of arbitrary n, you will pay anything for
> induction.
> >>
> >>Induction is your friend -- a magic bullet for many
> problems.
> >>
> >>Is induction natural for this problem? Well, just
> answer this
> >>question:
> >>
> >>How is the Gram-Schmidt process defined?
> >>
> >>Inductively, right?
> >>
> >>quasi
> >
> >On the other hand, if it fails for some n, for
> example n=2, then you
> >don't have to worry about induction.
> >
> >But my point was that if the result is true, then an
> inductive proof
> >is natural here since the Gram-Schmidt process is
> defined inductively.
> >If you can't clinch the induction, maybe check a few
> small values of n
> >such as n=1, n=2 to see if the result is even true.
> If the result is
> >false, then induction won't help you. Induction may
> be a magic bullet
> >but there's a limit to the magic -- it can't prove
> things true if
> >they're false.
> >
> >quasi
>
> One more comment about this.
>
> If you don't see a proof immediately, consider it
> automatic that you
> should then try to test it with some simple examples.
> It might lead to
> a counterexample, but if not, it may give you some
> intuition about why
> the result is true.
>
> Try a small special case where you actually carry out
> the Gram-Schmidt
> process, checking to see if relations v_i=w_i are
> forced.
>
> quasi




v1 =(3,0,4)
v2 =(-6,-4,1)
v3 =(5,0,-3)

These vectors are linearly indep but not orthogonal.
We will now apply Gram-Schmidt to get three vectors
w1,w2,w3 which span the same subspace (in this case,
all R^3) and orthogonal to each other.

First we take w1=v1=(3,0,4).
Now, w2= v2- ((w1*v2)/(||w1||^2))*w1
w2= (-108/25, -4, -81/25)

w3= v3- ((w1*v3)/(||w1||^2))*w1- ((w2*v3)/(||w2||^2))*w2

w3= ({1856/1129},{3132/1129},{1392/1129})

and so {w1,w2,w3} is an orthogonal set of vectors such
that {w1,w2,w3}={v1,v2,v3}.

But w1 does not equal v1??? They have different values!
I'm very confused as to how this example can claim what
i'm trying to prove, yet i don't see how what this
example says is true....
.

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