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From: GeometricGroup on 19 Jun 2010 19:32
In Michael Artin's Algebra page 243,

=======================================================
If W is a subspace of a vector space V, then the set of al vectors v which are orthogonal to every w in W is

W'={v \in V | <v, W>=0}.

This is a subspace of V and is called the orthogonal complement to W satisfying

V = W (+) W', where w \in W, <w,w> =/= 0 and W is a span of w such that W = {cw}.
========================================================

My question is,

Is there any restriction of a form <,> ? Any form, such as bilinear, hermitian, symmetric and quadratic form satisfies the above condition V = W (+) W', where W' is the orthogonal complement to W with respect to the given form?

For a given W, does every bilinear form (hemitian, symmetric form, etc) have the same W'?

Thanks.
From: Arturo Magidin on 20 Jun 2010 00:18
On Jun 19, 10:32 pm, GeometricGroup <ggx...(a)gmail.com> wrote:
> In Michael Artin's Algebra page 243,
>
> =======================================================
> If W is a subspace of a vector space V, then the set of al vectors v which are orthogonal to every w in W is
>
> W'={v \in V | <v, W>=0}.
>
> This is a subspace of V and is called the orthogonal complement to W satisfying
>
> V = W (+) W', where w \in W, <w,w> =/= 0 and W is a span of w such that W = {cw}.
> ========================================================

I don't understand the final sentence (the one that starts "where w in
W, <w,w>=/=-...") The rest seems perfectly sensible to me, but this
last part doesn't seem to be logically connected to what precedes it.
Particularly, "W is a span" seems like a very strange turn of
mathematical phrase.

(Then again, I don't have the book in front of me).

> My question is,
>
> Is there any restriction of a form <,> ? Any form, such as bilinear, hermitian, symmetric and quadratic form satisfies the above condition V = W (+) W', where W' is the orthogonal complement to W with respect to the given form?

Hmmm... Are you sure he isn't talking only about inner products?


> For a given W, does every bilinear form (hemitian, symmetric form, etc) have the same W'?

Different inner products may define different orthogonal complements.

For example, in the vector space of polynomial of degree at most 2
with real coefficients, with inner product

<p,q> = int_{-1}^1 p(x)q(x) dx

the vector x lies in the orthogonal complement of span(1,x^2).

But if we use the inner product int_{0}^1 p(x)q(x) dx instead, then

<1,x> = int_0^1 x dx = (1/2) =/= 0

so x is not in the orthogonal complement.

--
Arturo Magidin

From: David C. Ullrich on 21 Jun 2010 07:01
On Sat, 19 Jun 2010 23:32:36 EDT, GeometricGroup <ggx213(a)gmail.com>
wrote:

>In Michael Artin's Algebra page 243,
>
>=======================================================
>If W is a subspace of a vector space V, then the set of al vectors v which are orthogonal to every w in W is
>
>W'={v \in V | <v, W>=0}.
>
>This is a subspace of V and is called the orthogonal complement to W satisfying
>
>V = W (+) W', where w \in W, <w,w> =/= 0 and W is a span of w such that W = {cw}.
>========================================================

I don't believe the book says that. What does it really say,
_exactly_?

_Exactly_. Word for word, not changing a single letter, what does it
say?

(See, what you wrote doesn't quite make sense. So it's hard to
believe it's what Artin wrote.)

>
>My question is,
>
>Is there any restriction of a form <,> ? Any form, such as bilinear, hermitian, symmetric and quadratic form satisfies the above condition V = W (+) W', where W' is the orthogonal complement to W with respect to the given form?
>
>For a given W, does every bilinear form (hemitian, symmetric form, etc) have the same W'?
>
>Thanks.

From: GeometricGroup on 21 Jun 2010 06:23
> On Sat, 19 Jun 2010 23:32:36 EDT, GeometricGroup
> <ggx213(a)gmail.com>
> wrote:
>
> >In Michael Artin's Algebra page 243,
> >
> >=====================================================
> ==
> >If W is a subspace of a vector space V, then the set
> of al vectors v which are orthogonal to every w in W
> is
> >
> >W'={v \in V | <v, W>=0}.
> >
> >This is a subspace of V and is called the orthogonal
> complement to W satisfying
> >
> >V = W (+) W', where w \in W, <w,w> =/= 0 and W is a
> span of w such that W = {cw}.
> >=====================================================
> ===
>
> I don't believe the book says that. What does it
> really say,
> _exactly_?
>
> _Exactly_. Word for word, not changing a single
> letter, what does it
> say?
>
> (See, what you wrote doesn't quite make sense. So
> it's hard to
> believe it's what Artin wrote.)
>
> >
> >My question is,
> >
> >Is there any restriction of a form <,> ? Any form,
> such as bilinear, hermitian, symmetric and quadratic
> form satisfies the above condition V = W (+) W',
> where W' is the orthogonal complement to W with
> respect to the given form?
> >
> >For a given W, does every bilinear form (hemitian,
> symmetric form, etc) have the same W'?
> >
> >Thanks.
>

Michael Artin's Algebra p243,

====================================================
"

If W is a subspace of V, then we will denote by $W^{\bot}$ the set of all vectors v which are orthogonal to every $w \in W$:

(2.3) $W^{\bot}={v \in V | <v, W>=0},

This is a subspace of V, called the orthogonal complement to W.

(2.4) Proposition. Let $w \in V$ be a vector such that $<w, w> \neq = 0$. Let W={cw} be the span of w. Then V is the direct sum of W and its orthogonal complement:

$ V = W \oplus W^{\bot} $.

"
==================================================
From: Arturo Magidin on 21 Jun 2010 10:37
On Jun 21, 9:23 am, GeometricGroup <ggx...(a)gmail.com> wrote:
> > On Sat, 19 Jun 2010 23:32:36 EDT, GeometricGroup
> > <ggx...(a)gmail.com>
> > wrote:
>
> > >In Michael Artin's Algebra page 243,
>
> > >=====================================================
> > ==
> > >If W is a subspace of a vector space V, then the set
> > of al vectors v which are orthogonal to every w in W
> > is
>
> > >W'={v \in V | <v, W>=0}.
>
> > >This is a subspace of V and is called the orthogonal
> > complement to W satisfying
>
> > >V = W (+) W', where w \in W, <w,w> =/= 0 and W is a
> > span of w such that W = {cw}.
> > >=====================================================
> > ===
>
> > I don't believe the book says that. What does it
> > really say,
> > _exactly_?
>
> > _Exactly_. Word for word, not changing a single
> > letter, what does it
> > say?
>
> > (See, what you wrote doesn't quite make sense. So
> > it's hard to
> > believe it's what Artin wrote.)
>
> > >My question is,
>
> > >Is there any restriction of a form <,> ? Any form,
> > such as bilinear, hermitian, symmetric and quadratic
> > form satisfies the above condition V = W (+) W',
> > where W' is the orthogonal complement to W with
> > respect to the given form?
>
> > >For a given W, does every bilinear form (hemitian,
> > symmetric form, etc) have the same W'?
>
> > >Thanks.
>
> Michael Artin's Algebra p243,
>
> ====================================================
> "
>
> If W is a subspace of V, then we will denote by $W^{\bot}$ the set of all vectors v which are orthogonal to every $w \in W$:
>
> (2.3) $W^{\bot}={v \in V | <v, W>=0},
>
> This is a subspace of V, called the orthogonal complement to W.
>
> (2.4) Proposition. Let $w \in V$ be a vector such that $<w, w> \neq = 0$. Let W={cw} be the span of w. Then V is the direct sum of W and its orthogonal complement:
>
> $ V = W \oplus W^{\bot} $.
>
> "
> ==================================================

Which is very different from what you wrote before.

The key here is that in order to get that W and W^{\bot} intersect
trivially, you need <w,w> to be nonzero (if w is nonzero); otherwise,
you would have w, a nonzero vector, in both W and W^{\bot}. Artin is
not restricting himself to non-degenerate inner products (in which
<w,w> = 0 if and only if w=0). If you know that the inner product is
nondegenerate, then the proposition holds for more general subspaces
W, not just one dimensional ones.

--
Arturo Magidin
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